🔬 Model Lab

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D

I'll solve each problem step-by-step.

**Problem 1: Which runner finished fourth?**

Let me set up the relationships from the clues:
- Nico finished 11 minutes behind Pedro
- Olympia finished 3 minutes behind Pedro and 2 minutes ahead of Melina
- Olympia finished 6 minutes ahead of Luke

Let Pedro's time = 0 (reference point)
- Pedro: 0
- Olympia: 3 (3 min behind Pedro)
- Nico: 11 (11 min behind Pedro)
- Melina: 5 (2 min behind Olympia)
- Luke: 9 (6 min behind Olympia)

Ordering from fastest (lowest time) to slowest:
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

Luke finished fourth.

ANSWER 1: A

**Problem 2: Three friends with 6 identical pencils, each has at least one**

This is a stars and bars problem. If each of 3 friends gets at least 1 pencil, we distribute 6 - 3 = 3 remaining pencils among 3 friends with no restrictions.

The number of ways is C(3+3-1, 3-1) = C(5,2) = 10

ANSWER 2: D

**Problem 3: Rectangular garden to square garden**

Original rectangle: 50 ft × 10 ft
Perimeter = 2(50 + 10) = 120 ft
Original area = 50 × 10 = 500 sq ft

New square with same perimeter:
Perimeter = 120 ft
Side length = 120/4 = 30 ft
New area = 30 × 30 = 900 sq ft

Increase = 900 - 500 = 400 sq ft

ANSWER 3: D

**Problem 4: Gumdrops and replacement**

Total gumdrops: Green = 30, which is 25% of total
So total = 30/0.25 = 120 gumdrops

Blue: 30% × 120 = 36
Brown: 20% × 120 = 24

Half of blue gumdrops replaced with brown: 36/2 = 18 blue replaced
New brown count = 24 + 18 = 42

ANSWER 4: C

**Problem 5: Lowest score on one of two remaining tests**

First three test scores: 76 + 94 + 87 = 257
Average needed: 81 over 5 tests
Total needed: 81 × 5 = 405

Remaining points needed: 405 - 257 = 148

To minimize one test score, maximize the other.
Maximum on one test: 100
Minimum on the other: 148 - 100 = 48

ANSWER 5: A

**Problem 6: 4(299) + 3(299) + 2(299) + 298**

Factor out 299 where possible:
= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

**Problem 7: Reversed units and tens digits**

If a two-digit number has digits reversed, the difference depends on the digits.
If the original number is 10a + b (where a is tens, b is units), the reversed is 10b + a.

Difference = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)

The difference must be a multiple of 9. Looking at choices:
- 45 = 9 × 5 ✓
- 46: no
- 47: no
- 48: no
- 49: no

ANSWER 7: A

**Problem 8: Decimal point placement**

Without decimals: 75 × 256 = 19200
With decimals: 0.075 × 2.56

0.075 has 3 decimal places, 2.56 has 2 decimal places = 5 total
Result: 19200 with decimal moved 5 places left = 0.192

ANSWER 8: B

**Problem 9: 991 + 993 + 995 + 997 + 999 = 5000 - N**

Sum = 991 + 993 + 995 + 997 + 999
= (995 - 4) + (995 - 2) + 995 + (995 + 2) + (995 + 4)
= 5(995)
= 4975

So: 4975 = 5000 - N
N = 5000 - 4975 = 25

ANSWER 9: E

**Problem 10: (2 × 3 × 4)(1/2 + 1/3 + 1/4)**

= 24(1/2 + 1/3 + 1/4)
= 24(6/12 + 4/12 + 3/12)
= 24(13/12)
= 26

ANSWER 10: E

**Problem 11: Butter needed for 216 cookies**

One recipe: 15 cookies needs 3 tablespoons butter
For 216 cookies: 216/15 = 14.4 recipes

Since only full recipes: need 15 recipes (15 × 15 = 225 cookies)
Butter needed: 15 × 3 = 45 tablespoons
Sticks needed: 45/8 = 5.625, so round up to 6 sticks

ANSWER 11: B

**Problem 12: 5 times a number is 2**

5x = 2
x = 2/5

Reciprocal of x = 5/2
100 times the reciprocal = 100 × (5/2) = 250

ANSWER 12: D

Problem 1:

Let the runners’ finishing times be compared to Pedro’s time.

- Nico finished 11 minutes behind Pedro, so Nico is 11 minutes after Pedro.
- Olympia finished 3 minutes behind Pedro, so Olympia is 3 minutes after Pedro.
- Olympia finished 2 minutes ahead of Melina, so Melina is 2 minutes after Olympia.
- Olympia finished 6 minutes ahead of Luke, so Luke is 6 minutes after Olympia.

Now put them in order from earliest to latest, using Pedro as the reference:

- Pedro
- Olympia, 3 minutes later
- Melina, 2 minutes after Olympia, so 5 minutes later than Pedro
- Luke, 6 minutes after Olympia, so 9 minutes later than Pedro
- Nico, 11 minutes after Pedro

So the order is:

1. Pedro  
2. Olympia  
3. Melina  
4. Luke  
5. Nico

The runner who finished fourth is Luke.

ANSWER 1: A

Problem 2:

We need the number of ways to give 6 identical pencils to 3 friends, with each friend getting at least 1 pencil.

Let the numbers be positive integers \(x,y,z\) such that
\[
x+y+z=6.
\]

Count the positive solutions. This is a standard stars-and-bars problem:

The number of positive integer solutions to \(x+y+z=6\) is
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]

So there are 10 ways.

ANSWER 2: D

Problem 3:

The original garden is a rectangle 50 ft by 10 ft, so its area is
\[
50 \cdot 10 = 500 \text{ square feet}.
\]

Its perimeter is
\[
2(50+10)=2(60)=120 \text{ feet}.
\]

The same fence is used to make a square, so the square has perimeter 120 feet. Each side is
\[
120 \div 4 = 30 \text{ feet}.
\]

So the new square’s area is
\[
30 \cdot 30 = 900 \text{ square feet}.
\]

The increase in area is
\[
900-500=400.
\]

ANSWER 3: D

Problem 4:

Let the total number of gumdrops be \(T\).

We are told 30% are blue, 20% brown, 15% red, 10% yellow, and the remaining 30 are green.

The given percentages add to
\[
30+20+15+10=75\%.
\]
So green is the remaining
\[
25\%.
\]
And 25% corresponds to 30 gumdrops, so
\[
0.25T=30 \implies T=120.
\]

Now find the original numbers:

- Blue: \(30\%\) of 120 = 36
- Brown: \(20\%\) of 120 = 24

Half of the blue gumdrops are replaced with brown gumdrops, so:
- Blue decreases by 18
- Brown increases by 18

New number of brown gumdrops:
\[
24+18=42.
\]

ANSWER 4: C

Problem 5:

Shauna wants an average of 81 on 5 tests, so the total score must be
\[
5 \cdot 81 = 405.
\]

Her first three scores total
\[
76+94+87=257.
\]

So the last two tests must total
\[
405-257=148.
\]

We want the lowest score she could earn on one of the other two tests. Since test scores can be at most 100, to make one score as low as possible, make the other score as high as possible:
\[
148-100=48.
\]

So the lowest possible score is 48.

ANSWER 5: A

Problem 6:

Compute:
\[
4(299)+3(299)+2(299)+298.
\]

Factor the first three terms:
\[
(4+3+2)(299)+298=9(299)+298.
\]

Now calculate:
\[
9 \cdot 299 = 2691.
\]
Then
\[
2691+298=2989.
\]

ANSWER 6: B

Problem 7:

If one score has its tens digit and units digit reversed, suppose the original score is \(10a+b\) and the reversed score is \(10b+a\).

The difference is
\[
(10a+b)-(10b+a)=9(a-b).
\]

So the change in the total must be a multiple of 9.

Check the choices:
- 45 is not a multiple of 9
- 46 is not
- 47 is not
- 48 is not
- 49 is not

But the problem asks “might her incorrect sum have differed from the correct one?” Since the options are all near 45–49, we should look carefully: the difference in a reversed two-digit number is always a multiple of 9, and among the listed numbers, none are multiples of 9.

That means the intended choice must be the one that can occur as the absolute difference between sums from a single reversed score, which should be a multiple of 9. The only option that could fit a typical AMC-style intended answer is 45? No, 45 is not a multiple of 9? Actually 45 is a multiple of 9 because \(45=9\cdot 5\).

So the possible difference is 45.

ANSWER 7: A

Problem 8:

The correct product is
\[
0.075 \times 2.56.
\]

Compute:
\[
75 \times 256 = 19200.
\]
Since \(0.075\) has 3 decimal places and \(2.56\) has 2 decimal places, the product must have 5 decimal places total:
\[
19200 \to 0.19200 = 0.192.
\]

So the answer is \(0.192\).

ANSWER 8: B

Problem 9:

Add the numbers:
\[
991+993+995+997+999.
\]

These are 5 numbers centered at 995, and they form an arithmetic sequence. Their average is 995, so the sum is
\[
5 \cdot 995 = 4975.
\]

We are told
\[
4975 = 5000 - N.
\]
So
\[
N = 5000-4975 = 25.
\]

ANSWER 9: E

Problem 10:

Interpret the expression as
\[
(2\times 3\times 4)\left(\frac12+\frac13+\frac14\right).
\]

First compute:
\[
2\times 3\times 4 = 24.
\]

Now add the fractions:
\[
\frac12+\frac13+\frac14
= \frac{6}{12}+\frac{4}{12}+\frac{3}{12}
= \frac{13}{12}.
\]

Multiply:
\[
24 \cdot \frac{13}{12} = 2 \cdot 13 = 26.
\]

ANSWER 10: E

Problem 11:

Each recipe makes 15 cookies and uses 3 tablespoons of butter.

Walter and Gretel need 216 cookies, so the number of full recipes is
\[
216 \div 15 = 14.4.
\]
Since only full recipes are made, they need 15 full recipes.

Butter needed:
\[
15 \times 3 = 45 \text{ tablespoons}.
\]

There are 8 tablespoons in a stick, so
\[
45 \div 8 = 5.625.
\]
They need whole sticks, so they need 6 sticks.

ANSWER 11: B

Problem 12:

Let the number be \(x\). We are told
\[
5x=2,
\]
so
\[
x=\frac{2}{5}.
\]

The reciprocal of \(x\) is
\[
\frac{5}{2}.
\]

Now multiply by 100:
\[
100 \cdot \frac{5}{2} = 50 \cdot 5 = 250.
\]

ANSWER 12: D

Problem 1:  
We need to find who finished **fourth** out of Luke, Melina, Nico, Olympia, Pedro.

Given:
- **Nico** finished **11 minutes behind Pedro**  
  So: Nico = Pedro − 11 (Pedro is ahead)
- **Olympia** finished **2 minutes ahead of Melina**  
  So: Olympia = Melina + 2 (Olympia is ahead)
- **Olympia** finished **3 minutes behind Pedro**  
  So: Olympia = Pedro − 3
- **Olympia** finished **6 minutes ahead of Luke**  
  So: Olympia = Luke + 6 (Olympia is ahead)

Let Pedro’s time be \(P\). Then:
- Olympia = \(P - 3\)
- Nico = \(P - 11\)

Now Olympia is 6 minutes ahead of Luke:
\[
P - 3 = \text{Luke} + 6 \implies \text{Luke} = P - 9
\]

Also Olympia is 2 minutes ahead of Melina:
\[
P - 3 = \text{Melina} -? \text{(ahead means Olympia time is smaller)} 
\]
Interpret “Olympia finished 2 minutes ahead of Melina” as Olympia is **earlier**, i.e.
\[
\text{Melina} = \text{Olympia} + 2 = (P-3) + 2 = P-1
\]

So we have:
- Pedro: \(P\)
- Olympia: \(P-3\)
- Nico: \(P-11\)
- Luke: \(P-9\)
- Melina: \(P-1\)

Finishing order from **earliest (smallest time)** to **latest (largest time)**:
\[
P-11 \text{ (Nico)} < P-9 \text{ (Luke)} < P-3 \text{ (Olympia)} < P-1 \text{ (Melina)} < P \text{ (Pedro)}
\]

Thus the 4th place is **Melina**.

ANSWER 1: B


Problem 2:  
We distribute **6 identical pencils** among **3 friends**, each getting **at least 1** pencil.  
This is the number of positive integer solutions to:
\[
x+y+z=6,\quad x,y,z\ge 1
\]
Let:
\[
x'=x-1,\ y'=y-1,\ z'=z-1
\]
Then \(x',y',z'\ge 0\) and:
\[
(x'+1)+(y'+1)+(z'+1)=6 \implies x'+y'+z'=3
\]
Number of nonnegative integer solutions to \(x'+y'+z'=3\) is:
\[
\binom{3+3-1}{3}=\binom{5}{3}=10
\]

ANSWER 2: D


Problem 3:  
Rectangle: \(50 \times 10\).  
Area:
\[
50\cdot 10=500
\]

Same fence length means same perimeter.  
Rectangle perimeter:
\[
2(50+10)=2\cdot 60=120
\]
For a square, side length \(s\) satisfies:
\[
4s=120 \implies s=30
\]
Square area:
\[
30^2=900
\]
Enlargement:
\[
900-500=400
\]

ANSWER 3: D


Problem 4:  
Let total gumdrops be \(T\).  
Given:
- 30% blue \(\Rightarrow 0.30T\)
- 20% brown \(\Rightarrow 0.20T\)
- 15% red \(\Rightarrow 0.15T\)
- 10% yellow \(\Rightarrow 0.10T\)
- other 30 gumdrops are green \(\Rightarrow 30 = 0.30T\)

So:
\[
0.30T=30 \implies T=100
\]
Then:
- Blue = 30% of 100 = 30
- Brown initially = 20% of 100 = 20

If half of the blue are replaced with brown:
Half of 30 is 15 moved to brown.
So new brown:
\[
20+15=35
\]

ANSWER 4: A


Problem 5:  
Shauna wants an average of 81 on 5 tests.  
Total target score:
\[
81\cdot 5=405
\]
First three scores: \(76,94,87\). Sum:
\[
76+94=170,\quad 170+87=257
\]
So the other two tests must total:
\[
405-257=148
\]
To make one of the other two tests as low as possible, make the other as high as possible (max 100).  
So one lowest possible score is when the other is 100:
\[
\text{lowest} = 148-100=48
\]

ANSWER 5: A


Problem 6:  
Compute:
\[
4(299) + 3(299) + 2(299) + 298
\]
Combine like terms:
\[
(4+3+2)\cdot 299 + 298 = 9\cdot 299 + 298
\]
\[
9\cdot 299 = 9(300-1)=2700-9=2691
\]
Then:
\[
2691+298=2989
\]

ANSWER 6: B


Problem 7:  
Reversing the units and tens digits of one score changes the value by:
\[
\text{correct }=10a+b,\quad \text{wrong }=10b+a
\]
Difference:
\[
(10b+a)-(10a+b)=9(b-a)
\]
So the difference is a multiple of 9.

Check options: 45,46,47,48,49.  
Multiples of 9 near these:
- \(45=9\cdot 5\) yes
- \(46,47,48,49\) are not multiples of 9

So the difference could be 45.

ANSWER 7: A


Problem 8:  
Compute \(0.075 \times 2.56\) but calculator omitted decimals.

Let correct product be \(x\).  
If decimals were removed, each number is multiplied by the appropriate power of 10.

- \(0.075\) has 3 decimal places, so entering 75 instead multiplies by \(10^3\).
- \(2.56\) has 2 decimal places, so entering 256 instead multiplies by \(10^2\).

So the incorrect product shown is:
\[
75\cdot 256 = x\cdot 10^3 \cdot 10^2 = x\cdot 10^5
\]
Given incorrect result \(=19200\):
\[
19200 = x\cdot 10^5 \implies x = \frac{19200}{100000}=0.192
\]

ANSWER 8: B


Problem 9:  
Sum:
\[
991+993+995+997+999
\]
These are 5 consecutive odd numbers, average is the middle one:
Middle = 995.  
Sum = \(5 \times 995 = 4975\)

Given:
\[
991 + \cdots + 999 = 5000 - N
\]
So:
\[
5000-N=4975 \implies N=25
\]

ANSWER 9: E


Problem 10:  
\[
(2\cdot 3\cdot 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)
\]
First:
\[
2\cdot 3\cdot 4=24
\]
Now:
\[
\frac{1}{2}+\frac{1}{3}+\frac{1}{4}
\]
Common denominator 12:
\[
\frac{6}{12}+\frac{4}{12}+\frac{3}{12}=\frac{13}{12}
\]
Multiply:
\[
24\cdot \frac{13}{12} = 2\cdot 13=26
\]

ANSWER 10: E


Problem 11:  
Recipe makes 15 cookies using 3 tablespoons butter.  
So cookies per tablespoon:
\[
15 \text{ cookies} / 3 = 5 \text{ cookies per tablespoon}
\]
Walter and Gretel need 216 cookies.  
Required tablespoons:
\[
216/5 = 43.2
\]
But “only full recipes are made,” meaning butter comes in chunks of whole recipes.

Each full recipe makes 15 cookies.  
Number of full recipes needed:
\[
216/15 = 14.4
\]
So must use 15 recipes (since 14 would make 210, not enough).

Butter per recipe = 3 tablespoons:
\[
15\cdot 3 = 45 \text{ tablespoons}
\]
There are 8 tablespoons per stick:
\[
45/8 = 5.625
\]
Need a whole number of sticks, so 6 sticks.

ANSWER 11: B


Problem 12:  
“If 5 times a number is 2” means:
\[
5x=2 \implies x=\frac{2}{5}
\]
We want: “100 times the reciprocal of the number”:
Reciprocal of \(x\) is \(\frac{1}{x}=\frac{1}{2/5}=\frac{5}{2}\).
Then:
\[
100\cdot \frac{5}{2}=50\cdot 5=250
\]

ANSWER 12: D

Problem 1:
Let $P$ be the time Pedro finished. We represent the other runners' times relative to $P$:
- Nico: $N = P + 11$
- Olympia: $O = P + 3$
- Melina: $M = O + 2 = (P + 3) + 2 = P + 5$
- Luke: $L = O - 6 = (P + 3) - 6 = P - 3$
Ordering the runners by time (smaller time is faster):
1. Luke ($P - 3$)
2. Pedro ($P$)
3. Melina ($P + 5$)
4. Olympia ($P + 3$ is incorrect, let's re-evaluate: $P-3 < P < P+3 < P+5 < P+11$)
Wait, let's re-order:
Luke: $P-3$ (1st)
Pedro: $P$ (2nd)
Olympia: $P+3$ (3rd)
Melina: $P+5$ (4th)
Nico: $P+11$ (5th)
The runner who finished fourth is Melina.

ANSWER 1: B

Problem 2:
We need to find the number of ways to distribute 6 identical pencils to 3 friends such that each gets at least one. This is a "stars and bars" problem. Let $x_1 + x_2 + x_3 = 6$ where $x_i \ge 1$.
Let $y_i = x_i - 1$, so $y_1 + y_2 + y_3 = 6 - 3 = 3$ where $y_i \ge 0$.
The number of non-negative integer solutions is $\binom{n+k-1}{k-1} = \binom{3+3-1}{3-1} = \binom{5}{2} = 10$.

ANSWER 2: D

Problem 3:
Perimeter of the rectangle = $2(50 + 10) = 120$ feet.
The fence is used for a square, so each side of the square = $120 / 4 = 30$ feet.
Area of rectangle = $50 \times 10 = 500$ sq ft.
Area of square = $30 \times 30 = 900$ sq ft.
Increase = $900 - 500 = 400$ sq ft.

ANSWER 3: D

Problem 4:
Percentages: Blue 30%, Brown 20%, Red 15%, Yellow 10%. Total = 75%.
Green = 100% - 75% = 25%.
We are told 25% = 30 gumdrops. So, 1% = $30 / 25 = 1.2$ gumdrops.
Total gumdrops = $100 \times 1.2 = 120$.
Blue = $30\% \times 120 = 36$. Brown = $20\% \times 120 = 24$.
Half of blue (18) are replaced with brown:
New Brown = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
Total points needed for average of 81 over 5 tests = $81 \times 5 = 405$.
Sum of first three tests = $76 + 94 + 87 = 257$.
Remaining points needed = $405 - 257 = 148$.
To minimize one score, maximize the other (max score = 100).
Lowest score = $148 - 100 = 48$.

ANSWER 5: A

Problem 6:
$4(299) + 3(299) + 2(299) + 298 = (4+3+2)(299) + 298 = 9(299) + 298$.
$9(299) = 9(300 - 1) = 2700 - 9 = 2691$.
$2691 + 298 = 2989$.

ANSWER 6: B

Problem 7:
Let the number be $10t + u$. Reversing it gives $10u + t$.
The difference is $(10t + u) - (10u + t) = 9t - 9u = 9(t - u)$.
The difference must be a multiple of 9.
Among the choices, only 45 is a multiple of 9.

ANSWER 7: A

Problem 8:
$0.075$ has 3 decimal places, $2.56$ has 2. Total decimal places = 5.
The product 19200 must have the decimal moved 5 places to the left: $0.19200 = 0.192$.

ANSWER 8: B

Problem 9:
Sum = $(1000-9) + (1000-7) + (1000-5) + (1000-3) + (1000-1) = 5000 - (9+7+5+3+1) = 5000 - 25$.
Thus, $N = 25$.

ANSWER 9: E

Problem 10:
$(2 \times 3 \times 4) (\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = (24)(\frac{6}{12} + \frac{4}{12} + \frac{3}{12}) = 24(\frac{13}{12}) = 2 \times 13 = 26$.

ANSWER 10: E

Problem 11:
Recipes needed = $216 / 15 = 14.4$. Since only full recipes are made, they must make 15 recipes.
Butter needed = $15 \times 3 = 45$ tablespoons.
Sticks of butter = $45 / 8 = 5.625$.
Since they must supply the cookies, they need 6 sticks.

ANSWER 11: B

Problem 12:
$5x = 2 \implies x = 2/5$.
Reciprocal of $x = 5/2 = 2.5$.
$100 \times 2.5 = 250$.

ANSWER 12: D

Problem 1:
Five runners finished a race with the following time relations (relative to Pedro finishing at time 0): Nico finishes at +11, Olympia at +3, Melina at +5 (since Olympia finishes 2 minutes ahead of Melina), and Luke at +9 (since Olympia finishes 6 minutes ahead of Luke). Ordering by increasing finish time gives Pedro (1st), Olympia (2nd), Melina (3rd), Luke (4th), and Nico (5th). Thus Luke finished fourth.

ANSWER 1: A

Problem 2:
The number of ways to distribute 6 identical pencils to 3 friends with each getting at least one is the number of positive integer solutions to \(x + y + z = 6\). This equals \(\binom{5}{2} = 10\).

ANSWER 2: D

Problem 3:
The original rectangular garden has perimeter \(2(50 + 10) = 120\) feet. The square using the same fence has side length \(120/4 = 30\) feet and area \(900\) square feet. The original area is \(500\) square feet, so the increase is \(400\) square feet.

ANSWER 3: D

Problem 4:
The given percentages sum to 75%, so green gumdrops are the remaining 25% and total gumdrops \(= 30 / 0.25 = 120\). Blue gumdrops: \(0.3 \times 120 = 36\). Replacing half the blue gumdrops (18) with brown gumdrops increases the original brown count (\(0.2 \times 120 = 24\)) by 18, yielding 42 brown gumdrops.

ANSWER 4: C

Problem 5:
An average of 81 over five tests requires a total of \(5 \times 81 = 405\) points. The first three scores sum to \(76 + 94 + 87 = 257\), so the remaining two tests must total \(405 - 257 = 148\). To minimize one of these scores, maximize the other at 100; the minimum is then \(148 - 100 = 48\).

ANSWER 5: A

Problem 6:
Factor out the common term: \(4(299) + 3(299) + 2(299) + 298 = 9 \times 299 + 298 = 2691 + 298 = 2989\).

ANSWER 6: B

Problem 7:
Reversing the digits of a two-digit score \(10a + b\) changes the sum by \(|9(a - b)|\), a multiple of 9. Among the given options, only 45 is a multiple of 9 and therefore possible.

ANSWER 7: A

Problem 8:
The product \(0.075 \times 2.56\) has five decimal places in total. The calculator result 19200 therefore corresponds to the correctly placed decimal value 0.192.

ANSWER 8: B

Problem 9:
Rewrite each term: \(991 + 993 + 995 + 997 + 999 = 5 \times 1000 - (9 + 7 + 5 + 3 + 1) = 5000 - 25\). Thus \(N = 25\).

ANSWER 9: E

Problem 10:
First evaluate the sum inside the parentheses: \(\frac12 + \frac13 + \frac14 = \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{13}{12}\). Then multiply: \((2 \times 3 \times 4) \times \frac{13}{12} = 24 \times \frac{13}{12} = 26\).

ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies and uses 3 tablespoons of butter, so 216 cookies require \(216/15 = 14.4\) recipes. Only whole recipes may be made, so 15 recipes are required. Butter needed: \(15 \times 3 = 45\) tablespoons. With 8 tablespoons per stick, \(45/8 = 5.625\) sticks, so 6 sticks are needed.

ANSWER 11: B

Problem 12:
Solve \(5x = 2\) to obtain \(x = 2/5\). Then \(100/x = 100 \times (5/2) = 250\).

ANSWER 12: D

Problem 1:
The problem asks us to determine the order in which five runners finished a race, given certain time differences between their finish times, and identify who finished fourth.

Let's denote the finish time of each runner as follows: Luke (L), Melina (M), Nico (N), Olympia (O), and Pedro (P).

1. Nico finished 11 minutes behind Pedro: $N = P + 11$
2. Olympia finished 2 minutes ahead of Melina: $O = M + 2$
3. Olympia finished 3 minutes behind Pedro: $O = P - 3$
4. Olympia finished 6 minutes ahead of Luke: $O = L + 6$

From (3), we have $P - 3 = O$, and from (2), $M + 2 = O$. So, $P - 3 = M + 2$, giving $M = P - 5$.

From (3) and (4), $P - 3 = L + 6$, so $L = P - 9$.

Now, let's summarize the finish times relative to Pedro (P):
- $N = P + 11$
- $O = P - 3$
- $M = P - 5$
- $L = P - 9$

Ordering these from earliest to latest:
1. $L = P - 9$
2. $O = P - 3$
3. $M = P - 5$ (This is actually between $L$ and $O$, so the correct order for these three is $L, M, O$)
   - Correcting the order: $L (P - 9), M (P - 5), O (P - 3), P, N (P + 11)$
   - So, the correct order is: Luke, Melina, Olympia, Pedro, Nico.

Thus, Pedro finished fourth.

ANSWER 1: E

Problem 2:
The problem asks us to find the number of ways three friends can have a total of 6 identical pencils, with each friend having at least one pencil.

Let's denote the number of pencils each friend has as $a$, $b$, and $c$, where $a \geq 1$, $b \geq 1$, and $c \geq 1$. We are given that $a + b + c = 6$.

Since each friend must have at least one pencil, we can give one pencil to each friend first, leaving us with $6 - 3 = 3$ pencils to distribute among the three friends.

This is a classic stars and bars problem, where we have 3 stars (representing the 3 remaining pencils) and 2 bars (to divide among the 3 friends). The number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula $\binom{n + r - 1}{r - 1}$. Here, $n = 3$ and $r = 3$, so the number of ways is $\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}$.

Calculating $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Therefore, there are 10 ways to distribute the pencils.

ANSWER 2: D

Problem 3:
The problem asks us to find the increase in area when a rectangular garden with dimensions 50 feet by 10 feet is changed to a square garden using the same amount of fencing.

First, let's calculate the perimeter of the rectangular garden, which is $2 \times (50 + 10) = 120$ feet. This is the total length of the fence.

The perimeter of the square garden will also be 120 feet. Since a square has all sides equal, each side's length is $120 / 4 = 30$ feet.

The area of the original rectangular garden is $50 \times 10 = 500$ square feet.

The area of the square garden is $30 \times 30 = 900$ square feet.

The increase in area is $900 - 500 = 400$ square feet.

ANSWER 3: D

Problem 4:
The problem asks us to find the number of brown gumdrops after half of the blue gumdrops are replaced with brown ones in a jar containing gumdrops of different colors.

Given:
- 30% of the gumdrops are blue.
- 20% are brown.
- 15% are red.
- 10% are yellow.
- The remaining percentage is green, which equals 100% - (30% + 20% + 15% + 10%) = 25%.
- The number of green gumdrops is 30.

Since 25% of the gumdrops are green and this equals 30 gumdrops, we can find the total number of gumdrops. Let $T$ be the total number of gumdrops:
$0.25T = 30 \Rightarrow T = 30 / 0.25 = 120$.

So, there are 120 gumdrops in total.

Originally:
- Blue gumdrops = $0.30 \times 120 = 36$.
- Brown gumdrops = $0.20 \times 120 = 24$.

Half of the blue gumdrops are replaced with brown ones:
- Half of 36 blue gumdrops = $36 / 2 = 18$.
- So, 18 blue gumdrops are replaced with brown ones.

New number of brown gumdrops = original brown + 18 = $24 + 18 = 42$.

ANSWER 4: C

Problem 5:
The problem asks us to find the lowest score Shauna could earn on one of her last two tests to average 81 across five tests, given her scores on the first three tests.

Shauna's scores on the first three tests are 76, 94, and 87. The total score for these three tests is $76 + 94 + 87 = 257$.

To average 81 across 5 tests, the total score needed is $81 \times 5 = 405$.

Therefore, the total score needed for the last two tests is $405 - 257 = 148$.

To minimize one score, we need to maximize the other. The maximum score on a test is 100.

So, if one of the last two tests scores 100, the other test's score would be $148 - 100 = 48$.

Thus, the lowest score she could earn on one of the other two tests is 48.

ANSWER 5: A

Problem 6:
The problem asks us to simplify the expression $4(299) + 3(299) + 2(299) + 298$.

Let's simplify the given expression:
$4(299) + 3(299) + 2(299) + 298 = (4 + 3 + 2) \times 299 + 298$.
$= 9 \times 299 + 298$.
$= 299 \times 9 + 298$.
$= 299 \times (10 - 1) + 298$.
$= 2990 - 299 + 298$.
$= 2990 - 1$.
$= 2989$.

ANSWER 6: B

Problem 7:
The problem asks us to determine by how much Clara's incorrect sum might have differed from the correct one when she inadvertently reversed the units and tens digits of one score.

Let's consider a score $10a + b$, where $a$ is the tens digit and $b$ is the units digit. Reversing the digits gives $10b + a$.

The difference between the correct and incorrect score is $| (10a + b) - (10b + a) | = | 9a - 9b | = 9|a - b|$.

For $|a - b|$ to result in a difference that matches one of the given choices when multiplied by 9, let's examine the options:
- $9 \times 1 = 9$ (not an option)
- $9 \times 2 = 18$ (not an option)
- $9 \times 3 = 27$ (not an option)
- $9 \times 4 = 36$ (not an option, but $9 \times 5 = 45$ is an option, and $9 \times 6 = 54$ is not, but if we consider the absolute difference and the fact that we're looking for a possible difference, we should directly check the given options)
- Checking directly: $45 = 9 \times 5$, so if $|a - b| = 5$, the difference is 45.

Thus, a possible difference is indeed 45.

ANSWER 7: A

Problem 8:
The problem asks us to find

Problem 1:
Let Pedro's time be P. Nico finished 11 minutes behind Pedro: N = P + 11. Olympia finished 3 minutes behind Pedro: O = P + 3. Olympia finished 2 minutes ahead of Melina: O = M - 2 ⇒ M = O + 2 = P + 5. Olympia finished 6 minutes ahead of Luke: O = L - 6 ⇒ L = O + 6 = P + 9. The times in increasing order (fastest to slowest): Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11). The fourth finisher is Luke.
ANSWER 1: A

Problem 2:
We need the number of ways to distribute 6 identical pencils among 3 distinct friends, each getting at least 1. Let each friend take 1 pencil first, leaving 3 pencils to distribute freely among 3 people. The number of nonnegative integer solutions to y1 + y2 + y3 = 3 is C(3+3-1, 3-1) = C(5,2) = 10.
ANSWER 2: D

Problem 3:
Original garden: 50 ft × 10 ft, area = 500 sq ft. Perimeter = 2(50+10) = 120 ft. Square with same perimeter has side length 120/4 = 30 ft. Area of square = 30² = 900 sq ft. Enlargement = 900 - 500 = 400 sq ft.
ANSWER 3: D

Problem 4:
Percentages: blue 30%, brown 20%, red 15%, yellow 10%. Remaining green = 100% - 75% = 25% = 30 gumdrops. Total jar = 30 / 0.25 = 120 gumdrops. Blue = 0.30 × 120 = 36; brown = 0.20 × 120 = 24. Replacing half the blue (18) with brown increases brown to 24 + 18 = 42.
ANSWER 4: C

Problem 5:
Sum of first three tests: 76 + 94 + 87 = 257. Target total for five tests (average 81): 5 × 81 = 405. Needed sum for remaining two tests: 405 - 257 = 148. To minimize one test, maximize the other at 100; then the lowest possible score is 148 - 100 = 48.
ANSWER 5: A

Problem 6:
4(299) + 3(299) + 2(299) + 298 = (4+3+2)×299 + 298 = 9×299 + 298 = 2691 + 298 = 2989.
ANSWER 6: B

Problem 7:
Reversing the tens and units digits of a two-digit number changes its value by a multiple of 9 (difference = 9×(A-B) or 9×(B-A)). Among the choices, only 45 is a multiple of 9.
ANSWER 7: A

Problem 8:
Correct product 0.075 × 2.56. The product of the digits without decimals: 75 × 256 = 19200. Number of decimal places needed: 0.075 has 3, 2.56 has 2, total 5. Placing the decimal point: 0.19200 = 0.192.
ANSWER 8: B

Problem 9:
Sum = 991 + 993 + 995 + 997 + 999. The middle term is 995, so sum = 5 × 995 = 4975. Equation: 5000 - N = 4975 ⇒ N = 25.
ANSWER 9: E

Problem 10:
2 × 3 × 4 = 24. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12. Product = 24 × (13/12) = 2 × 13 = 26.
ANSWER 10: E

Problem 11:
Each recipe yields 15 cookies using 3 tablespoons butter. To get at least 216 cookies: 216/15 = 14.4, so 15 full recipes are needed. Total butter = 15 × 3 = 45 tablespoons. Number of sticks (8 tablespoons per stick) = 45/8 = 5.625. Whole sticks required: 6.
ANSWER 11: B

Problem 12:
5 times a number = 2 ⇒ the number = 2/5. The reciprocal = 5/2. One hundred times the reciprocal = 100 × (5/2) = 250.
ANSWER 12: D

# Problem 1

**Restate:** We need to determine the finishing order of five runners and identify who came in fourth.

**Set up:** Let Pedro's finishing time be $P$ minutes. "Behind" means a larger (slower) time; "ahead" means a smaller (faster) time.

- Nico finished 11 minutes behind Pedro: $\text{Nico} = P + 11$
- Olympia finished 3 minutes behind Pedro: $\text{Olympia} = P + 3$
- Olympia finished 2 minutes ahead of Melina: $\text{Melina} = \text{Olympia} + 2 = P + 5$
- Olympia finished 6 minutes ahead of Luke: $\text{Luke} = \text{Olympia} + 6 = P + 9$

**Compute the order** (smallest time = fastest = 1st place):

| Place | Runner | Time |
|-------|--------|------|
| 1st | Pedro | $P$ |
| 2nd | Olympia | $P + 3$ |
| 3rd | Melina | $P + 5$ |
| 4th | Luke | $P + 9$ |
| 5th | Nico | $P + 11$ |

**Rule out others:** Pedro is 1st, Olympia 2nd, Melina 3rd, Nico 5th — none of these are 4th.

ANSWER 1: A

---

# Problem 2

**Restate:** Find the number of ways to distribute 6 identical pencils among 3 friends so each gets at least 1.

**Set up:** Let the three friends receive $a, b, c$ pencils with $a, b, c \geq 1$ and $a + b + c = 6$.

Substitute $a' = a - 1,\; b' = b - 1,\; c' = c - 1$ so that $a', b', c' \geq 0$:
$$a' + b' + c' = 6 - 3 = 3$$

**Compute:** By stars and bars, the number of non-negative integer solutions is:
$$\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10$$

**Verify by listing:** The partitions of 6 into 3 positive parts are:
- (4, 1, 1) → 3 arrangements
- (3, 2, 1) → 6 arrangements
- (2, 2, 2) → 1 arrangement
- Total = 3 + 6 + 1 = 10 ✓

**Rule out others:** 1, 3, 6, and 12 all miss some arrangements.

ANSWER 2: D

---

# Problem 3

**Restate:** A 50 ft × 10 ft rectangular garden is re-fenced into a square using the same fence. Find the increase in area.

**Set up:**
- Perimeter of rectangle = $2(50 + 10) = 120$ feet
- Area of rectangle = $50 \times 10 = 500$ sq ft

The square uses the same 120 ft of fence:
- Side of square = $120 \div 4 = 30$ feet
- Area of square = $30 \times 30 = 900$ sq ft

**Compute:** Increase = $900 - 500 = 400$ sq ft.

**Rule out others:** 100, 200, 300, and 500 result from arithmetic errors (e.g., forgetting to double the perimeter or miscomputing the side length).

ANSWER 3: D

---

# Problem 4

**Restate:** Find the number of brown gumdrops after half the blue ones are replaced with brown ones.

**Set up:** The percentages for blue, brown, red, and yellow sum to:
$$30\% + 20\% + 15\% + 10\% = 75\%$$
So green gumdrops make up $100\% - 75\% = 25\%$ of the total.

Since 25% corresponds to 30 green gumdrops:
$$\text{Total} = \frac{30}{0.25} = 120 \text{ gumdrops}$$

**Compute:**
- Blue = $30\% \times 120 = 36$
- Brown = $20\% \times 120 = 24$

Half of the blue gumdrops: $36 \div 2 = 18$ are replaced with brown.

New brown count = $24 + 18 = 42$.

**Rule out others:** 35 and 36 come from not adding correctly; 48 and 64 come from replacing all blue or miscounting.

ANSWER 4: C

---

# Problem 5

**Restate:** Shauna's first three scores are 76, 94, 87. She needs an average of 81 over 5 tests (each max 100). Find the lowest possible score on one of the remaining two tests.

**Set up:**
$$\text{Total needed} = 81 \times 5 = 405$$
$$76 + 94 + 87 + x + y = 405$$
$$257 + x + y = 405 \implies x + y = 148$$

**Compute:** To minimize one score (say $x$), maximize the other ($y$). The maximum possible test score is 100:
$$x = 148 - 100 = 48$$

**Verify:** Scores 76, 94, 87, 48, 100 → sum = 405, average = 81. ✓

**Rule out others:** 52, 66, 70, 74 would correspond to not maximizing the other test score (e.g., assuming both remaining tests are equal gives 74, which is a tempting trap).

ANSWER 5: A

---

# Problem 6

**Restate:** Evaluate $4(299) + 3(299) + 2(299) + 298$.

**Set up:** Factor out 299 from the first three terms:
$$(4 + 3 + 2)(299) + 298 = 9(299) + 298$$

**Compute:**
$$9(299) = 9(300 - 1) = 2700 - 9 = 2691$$
$$2691 + 298 = 2989$$

**Rule out others:** 2889 comes from forgetting the 298 or miscounting; 2991 and 3009 come from sign errors in the $300 - 1$ step.

ANSWER 6: B

---

# Problem 7

**Restate:** Clara swapped the tens and units digits of one score. By how much could her total have changed?

**Set up:** Let the original score's last two digits be $a$ (tens) and $b$ (units). The original contribution is $10a + b$; the reversed contribution is $10b + a$.

**Compute the difference:**
$$|(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|$$

The error must be a **multiple of 9**.

**Check the choices:**
- 45 = 9 × 5 ✓ (e.g., digits 7 and 2: $72 \to 27$, difference = 45)
- 46, 47, 48, 49 are **not** multiples of 9 ✗

**Rule out others:** 46–49 are not divisible by 9, so they are impossible.

ANSWER 7: A

---

# Problem 8

**Restate:** Betty computed $0.075 \times 2.56$ but entered $75 \times 256$, getting 19200. Find the correct answer.

**Set up:** Count the total decimal places in the original numbers:
- $0.075$ has **3** decimal places
- $2.56$ has **2** decimal places
- Total = $3 + 2 = 5$ decimal places

**Compute:** Place the decimal point in 19200 so there are 5 digits after it:
$$19200 \to 0.19200 = 0.192$$

**Verify:** $0.075 \times 2.56 = \frac{75}{1000} \times \frac{256}{100} = \frac{19200}{100000} = 0.192$ ✓

**Rule out others:** .0192 has too many decimal places; 1.92, 19.2, and 192 have too few.

ANSWER 8: B

---

# Problem 9

**Restate:** Find $N$ given $991 + 993 + 995 + 997 + 999 = 5000 - N$.

**Set up:** Write each number as $1000 - k$:
$$991 = 1000 - 9$$
$$993 = 1000 - 7$$
$$995 = 1000 - 5$$
$$997 = 1000 - 3$$
$$999 = 1000 - 1$$

**Compute:**
$$\text{Sum} = 5(1000) - (9 + 7 + 5 + 3 + 1) = 5000 - 25$$

So $N = 25$.

**Rule out others:** 5, 10, 15, 20 come from incomplete or incorrect summation of the differences.

ANSWER 9: E

---

# Problem 10

**Restate:** Evaluate $(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$.

**Set up:** First compute the product: $2 \times 3 \times 4 = 24$.

Now distribute 24 across the 

 Problem 1:
We are asked to rank the five runners from fastest (first) to slowest (fifth) and identify who finished fourth.  
Let Pedro’s time be the baseline.  
- “Nico finished 11 minutes behind Pedro” → Nico = Pedro + 11.  
- “Olympia finished 3 minutes behind Pedro” → Olympia = Pedro + 3.  
- “Olympia finished 2 minutes ahead of Melina” → Melina = Olympia + 2 = Pedro + 5.  
- “Olympia finished 6 minutes ahead of Luke” → Luke = Olympia + 6 = Pedro + 9.  

Ordering by time (smaller is faster):  
1. Pedro (0)  
2. Olympia (+3)  
3. Melina (+5)  
4. Luke (+9)  
5. Nico (+11)  

Thus Luke finished fourth. The other choices occupy different positions: Pedro is 1st, Olympia 2nd, Melina 3rd, and Nico 5th.  
ANSWER 1: A

Problem 2:
We must count the number of ways to distribute 6 identical pencils to 3 friends so that each gets at least one.  
Let the friends receive \(x\), \(y\), and \(z\) pencils. Then \(x+y+z=6\) with \(x,y,z\ge 1\).  
Using stars and bars for positive integers, the count is  
\[
\binom{6-1}{3-1}=\binom{5}{2}=10.
\]  
We can also list the unordered partitions and their permutations:  
- (4,1,1): 3 permutations  
- (3,2,1): 6 permutations  
- (2,2,2): 1 permutation  
Total: \(3+6+1=10\).  
ANSWER 2: D

Problem 3:
We need the area increase when a 50 ft by 10 ft rectangle is reshaped into a square using the same fence.  
Original perimeter: \(2(50+10)=120\) feet.  
A square with perimeter 120 ft has side \(120/4=30\) ft.  
Original area: \(50\times 10=500\) sq ft.  
New area: \(30\times 30=900\) sq ft.  
Increase: \(900-500=400\) sq ft.  
ANSWER 3: D

Problem 4:
First find the total number of gumdrops.  
The given percentages sum to \(30+20+15+10=75\%\), so green gumdrops make up \(25\%\).  
If \(25\%\) of the total is 30, then the total is \(30/0.25=120\).  
Blue gumdrops: \(30\%\) of \(120 = 36\).  
Half of the blue gumdrops are \(36/2=18\). These are replaced by brown gumdrops.  
Original brown gumdrops: \(20\%\) of \(120 = 24\).  
New brown count: \(24+18=42\).  
ANSWER 4: C

Problem 5:
Shauna wants an average of 81 on 5 tests, so she needs a total of \(81\times 5 = 405\) points.  
Her first three scores sum to \(76+94+87=257\).  
Points needed on the last two tests: \(405-257=148\).  
To minimize one of these two scores, maximize the other. The maximum possible on one test is 100.  
Then the lowest possible score on the other test is \(148-100=48\).  
ANSWER 5: A

Problem 6:
Compute \(4(299)+3(299)+2(299)+298\).  
Combine the first three terms: \((4+3+2)(299)=9(299)\).  
Now \(9(299)=9(300-1)=2700-9=2691\).  
Add the remaining 298: \(2691+298=2989\).  
(Alternatively, \(9(299)+298=10(299)-1=2990-1=2989\).)  
ANSWER 6: B

Problem 7:
Let the correct score be \(10a+b\) (tens digit \(a\), units digit \(b\)).  
The reversed score is \(10b+a\).  
The difference between the incorrect sum and the correct sum is  
\[
|(10b+a)-(10a+b)| = |9b-9a| = 9|b-a|,
\]  
which must be a multiple of 9.  
Among the choices, only 45 is divisible by 9 (\(45=9\times 5\)).  
ANSWER 7: A

Problem 8:
Without decimal points Betty computed \(75\times 256=19200\).  
The correct factors are \(0.075\) and \(2.56\).  
\(0.075\) has 3 decimal places; \(2.56\) has 2 decimal places.  
Their product must have \(3+2=5\) decimal places.  
Placing the decimal point 5 places from the right in 19200 gives \(0.19200\), i.e., \(0.192\).  
ANSWER 8: B

Problem 9:
Compute the sum \(991+993+995+997+999\).  
Each term is 1000 minus a small number:  
\[
(1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1)=5000-25.
\]  
So the sum equals \(5000-25\).  
Given that this is \(5000-N\), we have \(N=25\).  
ANSWER 9: E

Problem 10:
Evaluate \((2\cdot 3\cdot 4)\left(\frac12+\frac13+\frac14\right)\).  
Distribute the product:  
\[
(2\cdot 3\cdot 4)\cdot\frac12 + (2\cdot 3\cdot 4)\cdot\frac13 + (2\cdot 3\cdot 4)\cdot\frac14
= (3\cdot 4) + (2\cdot 4) + (2\cdot 3)
= 12 + 8 + 6 = 26.
\]  
ANSWER 10: E

Problem 11:
Each full recipe makes 15 cookies and uses 3 tablespoons of butter.  
To supply at least 216 cookies using only full recipes:  
\(216\div 15 = 14.4\), so 14 recipes give only 210 cookies (not enough).  
They must make 15 recipes, producing 225 cookies.  
Butter needed: \(15\times 3 = 45\) tablespoons.  
Since there are 8 tablespoons per stick, \(45\div 8 = 5.625\).  
They must use whole sticks, so 6 sticks are required.  
ANSWER 11: B

Problem 12:
Let the number be \(x\).  
“5 times a number is 2” means \(5x=2\), so \(x=\frac25\).  
The reciprocal is \(\frac1x=\frac52\).  
Then 100 times the reciprocal is \(100\cdot\frac52 = 50\cdot 5 = 250\).  
ANSWER 12: D

Problem 1:
I need to determine the order of finish for all five runners.

Let me set Pedro's finish time as the reference point (time = 0).

- **Nico finished 11 minutes behind Pedro**: Nico's time = 0 + 11 = 11
- **Olympia finished 3 minutes behind Pedro**: Olympia's time = 0 + 3 = 3
- **Olympia finished 2 minutes ahead of Melina**: Melina's time = 3 + 2 = 5
- **Olympia finished 6 minutes ahead of Luke**: Luke's time = 3 + 6 = 9

Order from fastest (1st) to slowest (5th):
1. Pedro (0)
2. Olympia (3)
3. Melina (5)
4. Luke (9)
5. Nico (11)

The fourth-place finisher is Luke.

ANSWER 1: A

---

Problem 2:
I need to find the number of ways to distribute 6 identical pencils among 3 distinguishable friends, where each gets at least one pencil.

This is a stars-and-bars problem. I need the number of positive integer solutions to a + b + c = 6.

Using the formula C(n-1, k-1) where n = 6 and k = 3:
C(5, 2) = 10

I can verify by listing all partitions:
- 4+1+1: 3 arrangements
- 3+2+1: 6 arrangements  
- 2+2+2: 1 arrangement
Total = 3 + 6 + 1 = 10

ANSWER 2: D

---

Problem 3:
I need to find how much larger a square garden is compared to a rectangular garden with the same perimeter.

**Rectangular garden:** 50 ft × 10 ft
- Perimeter = 2(50 + 10) = 120 ft
- Area = 50 × 10 = 500 sq ft

**Square garden** with the same perimeter:
- Side length = 120 ÷ 4 = 30 ft
- Area = 30 × 30 = 900 sq ft

**Enlargement:** 900 - 500 = 400 sq ft

ANSWER 3: D

---

Problem 4:
I need to find the total number of gumdrops first, then calculate the new number of brown gumdrops.

Let T = total gumdrops.
- Blue: 30% = 0.30T
- Brown: 20% = 0.20T
- Red: 15% = 0.15T
- Yellow: 10% = 0.10T
- Green: 30

So: 0.30T + 0.20T + 0.15T + 0.10T + 30 = T
0.75T + 30 = T
30 = 0.25T
T = 120

Original amounts:
- Blue = 0.30 × 120 = 36
- Brown = 0.20 × 120 = 24

After replacing half the blue with brown:
- New brown = 24 + 36/2 = 24 + 18 = 42

ANSWER 4: C

---

Problem 5:
I need to find the lowest possible score on one of the remaining two tests.

**Total needed** for an average of 81 over 5 tests: 5 × 81 = 405
**Sum of first three tests:** 76 + 94 + 87 = 257
**Sum needed for last two tests:** 405 - 257 = 148

To minimize one score, I maximize the other. The maximum possible score is 100.
**Minimum score** = 148 - 100 = 48

ANSWER 5: A

---

Problem 6:
I need to compute 4(299) + 3(299) + 2(299) + 298.

= 299(4 + 3 + 2) + 298
= 299(9) + 298
= 2691 + 298
= 2989

ANSWER 6: B

---

Problem 7:
When the units and tens digits are swapped, the difference from the correct sum depends on the digit swap.

If a score has tens digit *a* and units digit *b*:
- Correct: 10a + b
- Reversed: 10b + a
- Difference: |(10a + b) - (10b + a)| = |9a - 9b| = 9|a - b|

The difference must be a multiple of 9. Checking the choices:
- 45 = 9 × 5 ✓
- 46: not divisible by 9
- 47: not divisible by 9
- 48: not divisible by 9
- 49: not divisible by 9

ANSWER 7: A

---

Problem 8:
I need to determine where the decimal point should go in 19200.

Without decimal points: 75 × 256 = 19200

The correct calculation: 0.075 × 2.56
- 0.075 has 3 decimal places
- 2.56 has 2 decimal places
- Total decimal places: 3 + 2 = 5

So the answer = 19200 ÷ 10⁵ = 0.19200 = 0.192

ANSWER 8: B

---

Problem 9:
I need to find N where 991 + 993 + 995 + 997 + 999 = 5000 - N.

Each term is 1000 minus a small number:
(1000 - 9) + (1000 - 7) + (1000 - 5) + (1000 - 3) + (1000 - 1)
= 5000 - (9 + 7 + 5 + 3 + 1)
= 5000 - 25

Therefore N = 25.

ANSWER 9: E

---

Problem 10:
I need to compute (2 × 3 × 4)(½ + ⅓ + ¼).

First part: 2 × 3 × 4 = 24

Second part: ½ + ⅓ + ¼ = 6/12 + 4/12 + 3/12 = 13/12

Product: 24 × 13/12 = 2 × 13 = 26

ANSWER 10: E

---

Problem 11:
I need to find how many sticks of butter are needed for 216 cookies.

One recipe makes 15 cookies using 3 tablespoons of butter.
Number of full recipes needed: ⌈216/15⌉ = ⌈14.4⌉ = 15 recipes

Butter needed: 15 × 3 = 45 tablespoons

Sticks of butter: ⌈45/8⌉ = ⌈5.625⌉ = 6 sticks

ANSWER 11: B

---

Problem 12:
I need to find 100 times the reciprocal of a number where 5 times the number equals 2.

5x = 2, so x = 2/5

Reciprocal of x = 5/2

100 × (5/2) = 250

ANSWER 12: D

Problem 1:
Let the finishing times (relative to a reference) be:
- Pedro: \(P\)
- Olympia is 3 min behind Pedro → Olympia: \(P+3\)
- Melina is 2 min behind Olympia → Melina: \((P+3)+2 = P+5\)
- Nico is 11 min behind Pedro → Nico: \(P+11\)
- Luke is 6 min behind Olympia → Luke: \((P+3)+6 = P+9\)

Thus the order (fastest to slowest) is:
1st Pedro, 2nd Olympia, 3rd Melina, **4th Luke**, 5th Nico.

ANSWER 1: A

---

Problem 2:
We need the number of positive integer solutions to  
\(a+b+c = 6\) where each \(a,b,c\ge1\).

Use stars‑and‑bars: place 2 separators among the 5 gaps between 6 stars:
\[
\binom{5}{2}=10.
\]

ANSWER 2: D

---

Problem 3:
Original rectangle perimeter = \(2(50+10)=120\) ft.  
For a square using the same fence, side = \(120/4=30\) ft, area = \(30^2=900\) ft².  
Original area = \(50\times10=500\) ft².  
Increase = \(900-500=400\) ft².

ANSWER 3: D

---

Problem 4:
Let total gumdrops be \(N\). Green = 30 gumdrops = 25 % of \(N\) → \(0.25N=30\) → \(N=120\).

Counts:
- Blue: \(0.30\times120=36\)
- Brown: \(0.20\times120=24\)
- Red: \(0.15\times120=18\)
- Yellow: \(0.10\times120=12\)
- Green: \(30\)

Half of the blue (18) are replaced with brown → new brown count = \(24+18=42\).

ANSWER 4: C

---

Problem 5:
Total points needed for average 81 on 5 tests = \(5\times81=405\).  
Points already earned = \(76+94+87=257\).  
Points remaining for two tests = \(405-257=148\).

To make one of those scores as low as possible, let the other test be 100 (max).  
Thus the lowest possible score = \(148-100=48\).

ANSWER 5: A

---

Problem 6:
\[
4(299)+3(299)+2(299)+298 = (4+3+2)\cdot299+298 = 9\cdot299+298.
\]
\(9\cdot299 = 9\cdot300-9 = 2700-9 = 2691\).  
\(2691+298 = 2989\).

ANSWER 6: B

---

Problem 7:
A two‑digit number with digits \(a\) (tens) and \(b\) (units) is \(10a+b\).  
Reversing gives \(10b+a\).  
Difference = \(|(10a+b)-(10b+a)| = |9a-9b| = 9|a-b|\).

Possible differences are multiples of 9: 0, 9, 18, 27, 36, 45, 54, 63, 72, 81.  
Among the answer choices only 45 is a multiple of 9.

ANSWER 7: A

---

Problem 8:
Betty entered the integers 75 (instead of 0.075) and 256 (instead of 2.56).  
\(75\times256=19200\).  

Correct product: \(0.075\times2.56 = \frac{75}{1000}\times\frac{256}{100}= \frac{19200}{100000}=0.192\).

ANSWER 8: B

---

Problem 9:
The five terms form an arithmetic sequence with average \(\frac{991+999}{2}=995\).  
Sum = \(5\times995=4975\) (or compute directly as 991+993+995+997+999 = 4985).  

Set \(991+993+995+997+999 = 5000 - N\).  
\(4985 = 5000 - N\) → \(N = 5000-4985 = 15\).

ANSWER 9: C

---

Problem 10:
\[
(2\times3\times4)\left(\frac12+\frac13+\frac14\right)=24\left(\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\right)
=24\left(\frac{13}{12}\right)=2\cdot13=26.
\]

ANSWER 10: E

---

Problem 11:
Each recipe makes 15 cookies and uses 3 tbsp butter.  
Need 216 cookies → at least \(\lceil 216/15\rceil = 15\) recipes.  

Butter required = \(15\times3=45\) tbsp.  
A stick holds 8 tbsp, so sticks needed = \(\lceil 45/8\rceil = 6\).

ANSWER 11: B

---

Problem 12:
\(5x=2\) → \(x=2/5\).  
Reciprocal = \(1/x = 5/2\).  
\(100\) times reciprocal = \(100\cdot5/2 = 250\).

ANSWER 12: D

Problem 1:
We are given the finishing times of five runners: Luke, Melina, Nico, Olympia, and Pedro. The goal is to determine who finished fourth.

- Nico finished 11 minutes behind Pedro: $ N = P + 11 $.
- Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro: $ O = P - 3 $ and $ M = O + 2 = P - 1 $.
- Olympia finished 6 minutes ahead of Luke: $ O = L + 6 $, so $ L = O - 6 = P - 9 $.

Now, we have the finishing times in terms of Pedro's time $ P $:
- Pedro: $ P $
- Olympia: $ P - 3 $
- Melina: $ P - 1 $
- Luke: $ P - 9 $
- Nico: $ P + 11 $

Order the runners by finishing time (lowest to highest):
1. Luke: $ P - 9 $
2. Olympia: $ P - 3 $
3. Melina: $ P - 1 $
4. Pedro: $ P $
5. Nico: $ P + 11 $

The fourth-place finisher is Pedro? No, wait: the order is Luke (1st), then Olympia (2nd), then Melina (3rd), then Pedro (4th), then Nico (5th). Thus, the fourth-place finisher is Pedro? No, Pedro is 4th, but the question asks for fourth, so Pedro is 4th? Wait, no: the order is Luke, Olympia, Melina, Pedro, Nico. So fourth is Pedro? But the choices include Pedro, but let's see: the fourth runner is Pedro? Actually, the order is:

1. Luke
2. Olympia
3. Melina
4. Pedro
5. Nico

So fourth is Pedro? But the choices are A. Luke, B. Melina, C. Nico, D. Olympia, E. Pedro. So fourth is Pedro, choice E.

But wait, is that right? Let me recheck: the times are:
- Luke: P - 9
- Olympia: P - 3
- Melina: P - 1
- Pedro: P
- Nico: P + 11

So order from earliest to latest: Luke (P-9), then Olympia (P-3), then Melina (P-1), then Pedro (P), then Nico (P+11). So fourth is Pedro.

But the question is "which runner finished fourth?" So Pedro is fourth.

But let me confirm: is there any mistake? Olympia is 3 minutes behind Pedro, so Pedro finishes after Olympia. Melina is 2 minutes behind Olympia, so Melina is after Olympia but before Pedro. Luke is 6 minutes behind Olympia, so Luke is first. Nico is 11 minutes behind Pedro, so last. So order: Luke, Olympia, Melina, Pedro, Nico. Fourth is Pedro.

ANSWER 1: E

Problem 2:
Three friends have 6 identical pencils, each with at least one pencil. We need to find the number of ways to distribute the pencils.

This is a stars and bars problem with each friend getting at least one pencil. Let the friends be A, B, C. We need to find the number of solutions to $ a + b + c = 6 $ with $ a, b, c \geq 1 $.

Let $ a' = a - 1 $, $ b' = b - 1 $, $ c' = c - 1 $, then $ a' + b' + c' = 3 $ with $ a', b', c' \geq 0 $.

Number of non-negative integer solutions is $ \binom{3 + 3 - 1}{3} = \binom{5}{3} = 10 $.

The choices are 1, 3, 6, 10, 12. So 10 is choice D.

We can list:
(1,1,4), (1,4,1), (4,1,1)
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
(2,2,2)

That's 3 + 6 + 1 = 10.

ANSWER 2: D

Problem 3:
A rectangular garden is 50 feet long and 10 feet wide, so area is $ 50 \times 10 = 500 $ square feet. The perimeter is $ 2 \times (50 + 10) = 120 $ feet.

When changed to a square with the same perimeter, each side is $ 120 / 4 = 30 $ feet. Area is $ 30 \times 30 = 900 $ square feet.

Increase in area: $ 900 - 500 = 400 $ square feet.

Choices: 100, 200, 300, 400, 500. So 400 is D.

ANSWER 3: D

Problem 4:
A jar has gumdrops: 30% blue, 20% brown, 15% red, 10% yellow, and 30 green. Green is 100% - (30+20+15+10)% = 25%? No, the percentages add to 30+20+15+10=75%, so green is 25%? But it says "the other 30 gumdrops are green", so green is 30 gumdrops.

Let total gumdrops be T. Then green is 25% of T? But 25% of T = 30? No, the percentages: blue 30%, brown 20%, red 15%, yellow 10%, so together 75%, so green is 25%, and 25% of T = 30, so T = 30 / 0.25 = 120.

So total gumdrops: 120.

Blue: 30% of 120 = 36
Brown: 20% of 120 = 24
Red: 15% of 120 = 18
Yellow: 10% of 120 = 12
Green: 30

Half of blue gumdrops are replaced with brown. Half of blue: 36 / 2 = 18. So blue decreases by 18, brown increases by 18.

New brown: 24 + 18 = 42.

Choices: 35,36,42,48,64. So 42 is C.

ANSWER 4: C

Problem 5:
Shauna has five tests. First three: 76, 94, 87. Sum so far: 76 + 94 = 170, +87 = 257.

She wants average 81 for five tests, so total sum needed: 5 * 81 = 405.

Sum of last two tests needed: 405 - 257 = 148.

To find the lowest score on one of the other two tests, we maximize the other score. Maximum is 100, so if one is 100, the other is 148 - 100 = 48.

Choices: 48,52,66,70,74. So 48 is A.

ANSWER 5: A

Problem 6:
Compute $ 4 \times 299 + 3 \times 299 + 2 \times 299 + 298 $.

Factor 299: $ (4+3+2) \times 299 + 298 = 9 \times 299 + 298 $.

Now, 9*299 = 9*(300 - 1) = 2700 - 9 = 2691.

Then 2691 + 298 = 2989.

Choices: 2889,2989,2991,2999,3009. So 2989 is B.

ANSWER 6: B

Problem 7:
Clara reversed the units and tens digit of one score. The difference between correct and incorrect sum.

Suppose the original number is 10a + b, and she reversed to 10b + a. The difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b).

So the difference is a multiple of 9.

Choices: 45,46,47,48,49. Which is multiple of 9? 45 is 9*5, so 45.

The incorrect sum could be higher or lower, but the difference magnitude is multiple of 9. So 45 is possible.

ANSWER 7: A

Problem 8:
Betty computed 0.075 × 2.56 but forgot decimals, so she did 75 × 256 = 19200.

Now, 0.075 has three decimal places? 0.075 is 75/1000, 2.56 is 256/100, so product is (75/1000) * (256/100) = (75*256) / 100000.

75*256 = 19200, so 19200 / 100000 = 0.192.

Choices: 0.0192, 0.192, 1.92, 19.2, 192. So 0.192 is B.

ANSWER 8: B

Problem 9:
$ 991 + 993 + 995 + 997 + 999 = 5000 - N $

Compute the sum: these are five consecutive odd numbers, average is 995, so sum = 5 * 995 = 4975.

991 is 995 - 4, 993 is 995 - 2, 995, 997 is 995 + 2, 999 is 995 + 4. So sum = 5*995 = 4975.

So 4975 = 5000 - N, thus N = 5000 - 4975 = 25.

Choices: 5,10,15,20,25. So 25 is E.

ANSWER 9: E

Problem 10:
$ (2 \times 3 \times 4) \times \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) $

First, 2*3*4 = 24.

Now, $ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6}{12} + \frac{4}{

### Problem 1:
What is asked: We need to order 5 runners by their finish times to find who placed 4th. We are given all relative finish times, so we assign Pedro (a common reference point) a finish time of 0 (smaller numbers = faster, earlier finish):
- Nico is 11 minutes behind Pedro: Nico = 0 + 11 = 11
- Olympia is 3 minutes behind Pedro: Olympia = 0 + 3 = 3
- Olympia is 2 minutes ahead of Melina: Melina = 3 + 2 = 5
- Olympia is 6 minutes ahead of Luke: Luke = 3 + 6 = 9

Order from first to last place: 1. Pedro (0), 2. Olympia (3), 3. Melina (5), 4. Luke (9), 5. Nico (11). The 4th place runner is Luke. Tempting wrong answers: Melina (if you mix up who is ahead/behind) or Nico (if you reverse the finish order).
ANSWER 1: A

---
### Problem 2:
What is asked: Count the number of ways to split 6 identical pencils among 3 distinct friends, each with at least 1 pencil. This is equivalent to counting positive integer solutions to \(x+y+z=6\). We list all valid ordered triples (friends are distinct, so order matters):
- (1,1,4) and permutations: 3 ways
- (1,2,3) and permutations: 6 ways
- (2,2,2): 1 way
Total = 3+6+1=10. A common wrong answer is 3 (only counting unordered partitions, ignoring that friends are distinct).
ANSWER 2: D

---
### Problem 3:
What is asked: Calculate the increase in area when a rectangular garden is converted to a square using the same total fence length. First calculate the original perimeter of the rectangle: \(2*(50+10)=120\) ft. The square has the same perimeter, so each side is \(120/4=30\) ft. Original area: \(50*10=500\) sq ft. Square area: \(30*30=900\) sq ft. The increase is \(900-500=400\) sq ft.
ANSWER 3: D

---
### Problem 4:
What is asked: Find the new number of brown gumdrops after half the blue gumdrops are converted to brown. First sum the given percentages: 30%+20%+15%+10%=75%, so green gumdrops are 25% of the total, equal to 30. Total gumdrops = \(30/0.25=120\). Original blue: \(0.3*120=36\), original brown: \(0.2*120=24\). Add half the blue gumdrops (18) to brown: \(24+18=42\).
ANSWER 4: C

---
### Problem 5:
What is asked: Find the lowest possible score Shauna can get on one of her last two tests to average 81 across all 5 tests. Total points needed for an 81 average: \(5*81=405\). Sum of her first 3 tests: \(76+94+87=257\). The sum of the last two tests must be \(405-257=148\). To minimize one score, maximize the other (max 100 points), so the lowest possible score is \(148-100=48\).
ANSWER 5: A

---
### Problem 6:
What is asked: Simplify the expression \(4(299)+3(299)+2(299)+298\). Factor out 299 from the first three terms: \((4+3+2)*299 + 298 = 9*299 + 298\). Rewrite 298 as \(299-1\): \(10*299 -1 = 2990-1=2989\).
ANSWER 6: B

---
### Problem 7:
What is asked: Identify which difference could result from reversing the tens and units digits of a number. A 2-digit number is \(10a+b\); reversed it is \(10b+a\). The difference is \(9(b-a)\), so the difference must be a multiple of 9. Only 45 (9*5) is a multiple of 9.
ANSWER 7: A

---
### Problem 8:
What is asked: Find the correct value of \(0.075*2.56\) when the integer product 75*256=19200. 0.075 has 3 decimal places, 2.56 has 2, so total 5 decimal places. Move the decimal in 19200 five places left: \(19200/100000=0.192\).
ANSWER 8: B

---
### Problem 9:
What is asked: Solve for N in \(991+993+995+997+999=5000-N\). Rewrite each term as \(1000 - k\): sum = \(5*1000 - (9+7+5+3+1)=5000-25\), so N=25.
ANSWER 9: E

---
### Problem 10:
What is asked: Simplify \((2×3×4)(1/2 + 1/3 + 1/4)\). The product 2*3*4=24. Distribute 24: \(24*(1/2)+24*(1/3)+24*(1/4)=12+8+6=26\).
ANSWER 10: E

---
### Problem 11:
What is asked: Calculate the number of butter sticks needed to make at least 216 cookies. Each full recipe makes 15 cookies, so number of recipes = ceiling(216/15)=15 (14 recipes only make 210, which is too few). Total butter needed: 15*3=45 tbsp. Number of sticks = ceiling(45/8)=6 (5 sticks only have 40 tbsp, too few).
ANSWER 11: B

---
### Problem 12:
What is asked: If 5 times a number is 2, find 100 times the reciprocal of the number. Let the number be x: 5x=2 → x=2/5. Reciprocal of x is 5/2=2.5. 100*2.5=250.
ANSWER 12: D

### Problem 1:
We need to rank the five runners by finish time (lower time = better finish). Let Pedro’s time be \( P \).
- Nico finished 11 minutes behind Pedro: Nico’s time = \( P + 11 \).
- Olympia finished 3 minutes behind Pedro: Olympia’s time = \( P + 3 \).
- Olympia finished 2 minutes ahead of Melina: Melina is 2 minutes behind Olympia, so Melina’s time = \( (P+3) + 2 = P +5 \).
- Olympia finished 6 minutes ahead of Luke: Luke is 6 minutes behind Olympia, so Luke’s time = \( (P+3) +6 = P +9 \).
Ranking from fastest to slowest:
1.  Pedro (\( P \))
2.  Olympia (\( P+3 \))
3.  Melina (\( P+5 \))
4.  Luke (\( P+9 \))
5.  Nico (\( P+11 \))
The fourth-place finisher is Luke.
ANSWER 1: A
---
### Problem 2:
We need the number of ways to distribute 6 identical pencils to 3 distinct friends, with each friend getting at least 1 pencil. This is a stars and bars problem for positive integer solutions to \( x + y + z = 6 \), where \( x,y,z \) are the number of pencils each friend has.
Let \( x' = x-1 \), \( y' = y-1 \), \( z' = z-1 \), so \( x',y',z' \geq 0 \) and \( x' + y' + z' = 6-3 = 3 \). The number of non-negative solutions is \( \binom{3 + 3 -1}{3-1} = \binom{5}{2} = 10 \).
We can also verify by listing partitions:
- \( 4,1,1 \): 3 permutations (which friend gets 4)
- \( 3,2,1 \): \( 3! =6 \) permutations (all distinct values)
- \( 2,2,2 \): 1 permutation (all equal)
Total: \( 3+6+1=10 \).
ANSWER 2: D
---
### Problem 3:
First calculate the perimeter of the original rectangular garden, which equals the length of the fence:
Perimeter = \( 2*(length + width) = 2*(50 +10) = 120 \) feet.
For the square garden with the same perimeter, each side length is \( 120 /4 = 30 \) feet.
Original area: \( 50*10 = 500 \) square feet.
New square area: \( 30*30 = 900 \) square feet.
Enlargement: \( 900 - 500 = 400 \) square feet.
ANSWER 3: D
---
### Problem 4:
First find the total number of gumdrops. The given percentages add to \( 30\% +20\% +15\% +10\% =75\% \), so green gumdrops make up \( 100\% -75\% =25\% \). We know 25% of the total is 30 gumdrops, so total gumdrops \( T = 30 / 0.25 = 120 \).
Original counts:
- Blue: \( 30\% *120 = 36 \)
- Brown: \( 20\% *120 =24 \)
Half the blue gumdrops are replaced with brown: half of 36 is 18. So we remove 18 blue and add 18 brown.
New brown count: \( 24 + 18 =42 \).
ANSWER 4: C
---
### Problem 5:
To average 81 over 5 tests, the total score needed is \( 81 *5 =405 \).
Sum of the first three tests: \(76 +94 +87 =257 \).
Total needed for the last two tests: \(405 -257 =148 \).
To minimize the score of one test, maximize the other (maximum possible score is 100). So the lowest possible score is \(148 -100 =48 \).
Verify: \(76 +94 +87 +100 +48 =405\), average \(405/5=81\), which works.
ANSWER 5: A
---
### Problem 6:
Factor out 299 from the first three terms:
\(4(299) +3(299) +2(299) +298 = (4+3+2)*299 +298 =9*299 +298\).
Note \(299 = 300 -1\), so:
\(9*(300 -1) +298 = 2700 -9 +298 = 2700 + 289 =2989\).
ANSWER 6: B
---
### Problem 7:
Let the original score be a two-digit number \(10a + b\) (a = tens digit, b = units digit). Reversing the digits gives \(10b +a\). The difference between the correct and incorrect sum is:
\(|(10a +b) - (10b +a)| = |9a -9b| =9|a -b|\).
This difference must be a multiple of 9. Among the choices, only 45 is divisible by 9 (\(45=9*5\), e.g., original score 61 reversed to 16 gives a difference of 45).
ANSWER 7: A
---
### Problem 8:
Betty entered the numbers without decimals: \(075 *256 =75*256=19200\), which matches the calculator result.
The original numbers have \(3 + 2 =5\) total decimal places, so the correct product is 19200 with the decimal moved 5 places left: \(0.192\).
Verify directly: \(0.075 *2.56 =0.192\).
ANSWER 8: B
---
### Problem 9:
Rewrite each term as \(1000 - k\):
\(991=1000-9\), \(993=1000-7\), \(995=1000-5\), \(997=1000-3\), \(999=1000-1\).
Sum: \(5*1000 - (9+7+5+3+1) =5000 -25 =4975\).
We have \(4975 =5000 -N\), so \(N=25\).
ANSWER 9: E
---
### Problem 10:
First calculate \(2*3*4=24\). The expression is \(24*(\frac{1}{2} + \frac{1}{3} + \frac{1}{4})\).
Distribute the 24:
\(24*\frac{1}{2} +24*\frac{1}{3} +24*\frac{1}{4} =12 +8 +6=26\).
ANSWER 10: E
---
### Problem 11:
Each recipe makes 15 cookies, so the number of full recipes needed for at least 216 cookies is \( \lceil 216/15 \rceil =15 \) (since 14 recipes make 210 cookies, which is too few).
Total butter needed: \(15 *3 =45\) tablespoons.
There are 8 tablespoons per stick, so number of sticks is \( \lceil 45/8 \rceil =6 \) (5 sticks give 40 tablespoons, which is too little).
ANSWER 11: B
---
### Problem 12:
Let the number be \(x\). We know \(5x=2\), so \(x=2/5\). The reciprocal of \(x\) is \(1/x =5/2=2.5\).
100 times the reciprocal: \(100 *2.5=250\).
ANSWER 12: D