🔬 Model Lab

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

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Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

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Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

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Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B

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

**Problem 1:**

We need to order: 15/11, 19/15, and 17/13 from least to greatest.

Let me convert each to a decimal or use a common approach:
- 15/11 = 1.3636...
- 19/15 = 1.2666...
- 17/13 = 1.3076...

Ordering from least to greatest: 19/15 < 17/13 < 15/11

That corresponds to: 19/15 < 17/13 < 15/11

Looking at the choices, this matches option E.

ANSWER 1: E

---

**Problem 2:**

Using the principle of inclusion-exclusion:
- Total students = 39
- Students with dog = 20
- Students with cat = 26
- Students with both = ?

Formula: Total = (dog) + (cat) − (both)

39 = 20 + 26 − (both)
39 = 46 − (both)
(both) = 46 − 39 = 7

ANSWER 2: A

---

**Problem 3:**

Starting investment: $100

After first year (15% loss):
Remaining = 100 × (1 − 0.15) = 100 × 0.85 = $85

After second year (20% gain):
Final = 85 × (1 + 0.20) = 85 × 1.20 = $102

Overall change: 102 − 100 = $2 gain
Percentage change: 2/100 = 2% gain

ANSWER 3: D

---

**Problem 4:**

Wind chill = (air temperature) − 0.7 × (wind speed)
Wind chill = 36 − 0.7 × 18
Wind chill = 36 − 12.6
Wind chill = 23.4 °F

Closest to 23.

ANSWER 4: B

---

**Problem 5:**

Let total marbles = n
- All but 6 are red: red = n − 6
- All but 8 are green: green = n − 8
- All but 4 are blue: blue = n − 4

Since these account for all marbles:
(n − 6) + (n − 8) + (n − 4) = n
3n − 18 = n
2n = 18
n = 9

ANSWER 5: C

---

**Problem 6:**

Goal: Average of 95 on four 100-point tests.
Total needed: 95 × 4 = 380 points

First two tests: 97 + 91 = 188
Remaining two tests need: 380 − 188 = 192 points

For the third test to allow reaching the goal, after taking it, she must still be able to reach 380 total with the fourth test (max 100).

Let third test score = x
Then: 188 + x + (fourth test) = 380
Fourth test = 192 − x

For this to be possible: 192 − x ≤ 100
Therefore: x ≥ 92

The lowest possible score on the third test is 92.

ANSWER 6: B

---

**Problem 7:**

Let the heights be 3k and 4k (in ratio 3:4, with the taller being 4k).

Given: 4k − 3k = 16
k = 16

Taller tree height: 4k = 4(16) = 64 feet

ANSWER 7: B

---

**Problem 8:**

In a single-elimination tournament, each game eliminates exactly one team.
Starting with 16 teams, we need to eliminate 15 teams to get 1 winner.
Therefore, 15 games are needed.

ANSWER 8: D

---

**Problem 9:**

Find all ways to write 100 = a × b × c where a < b < c and all are positive integers.

100 = 2² × 5²

Factorizations with a < b < c:
- 1 × 4 × 25: ✓
- 1 × 5 × 20: ✓
- 1 × 10 × 10: ✗ (b = c)
- 2 × 2 × 25: ✗ (a = b)
- 2 × 5 × 10: ✓
- 4 × 5 × 5: ✗ (b = c)

Valid choices: (1,4,25), (1,5,20), (2,5,10)

That's 3 ways.

ANSWER 9: D

---

**Problem 10:**

Three-digit numbers: 100 to 999

Smallest three-digit multiple of 13: ⌈100/13⌉ × 13 = 8 × 13 = 104
Largest three-digit multiple of 13: ⌊999/13⌋ × 13 = 76 × 13 = 988

Number of multiples: 76 − 8 + 1 = 69

ANSWER 10: C

---

**Problem 11:**

To maximize the difference, subtract the smallest from the largest:
Largest: 12
Smallest: −16

Difference: 12 − (−16) = 12 + 16 = 28

ANSWER 11: D

---

**Problem 12:**

As a boy: 15 miles in 3.5 hours = 15/3.5 miles per hour
Time per mile = 3.5/15 hours = 210/15 minutes = 14 minutes per mile

As an old man: 10 miles in 4 hours
Time per mile = 4/10 hours = 24 minutes per mile

Difference: 24 − 14 = 10 minutes longer

ANSWER 12: B

Problem 1:
We need to compare the fractions
\[
\frac{15}{11},\quad \frac{19}{15},\quad \frac{17}{13}
\]
from least to greatest.

A quick way is to approximate each one:

- \(\frac{15}{11} \approx 1.3636\)
- \(\frac{19}{15} \approx 1.2667\)
- \(\frac{17}{13} \approx 1.3077\)

So the order from least to greatest is
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]

That matches choice E.

ANSWER 1: E

Problem 2:
Each student has at least one pet: a dog, a cat, or both.

Let:
- \(D = 20\) students with a dog
- \(C = 26\) students with a cat
- Total students \(= 39\)

If \(x\) students have both, then by inclusion-exclusion:
\[
20 + 26 - x = 39.
\]
So
\[
46 - x = 39
\]
\[
x = 7.
\]

Therefore, 7 students have both a dog and a cat.

ANSWER 2: A

Problem 3:
Tammy starts with \$100.

First year: 15% loss.
\[
100 \times 0.85 = 85
\]

Second year: 20% gain on the remaining \$85.
\[
85 \times 1.20 = 102
\]

So her investment went from \$100 to \$102, which is a gain of \$2, or 2%.

ANSWER 3: D

Problem 4:
Use the formula:
\[
\text{wind chill} = \text{air temperature} - 0.7(\text{wind speed})
\]

Substitute \(36^\circ\text{F}\) and 18 mph:
\[
36 - 0.7(18) = 36 - 12.6 = 23.4
\]

The closest choice is 23.

ANSWER 4: B

Problem 5:
Let the total number of marbles be \(N\).

- “All but 6 are red” means red marbles \(= N - 6\)
- “All but 8 are green” means green marbles \(= N - 8\)
- “All but 4 are blue” means blue marbles \(= N - 4\)

Since every marble is either red, green, or blue, the counts must add to \(N\):
\[
(N-6) + (N-8) + (N-4) = N
\]
\[
3N - 18 = N
\]
\[
2N = 18
\]
\[
N = 9
\]

So there are 9 marbles total.

ANSWER 5: C

Problem 6:
She wants an average of 95 on 4 tests, so the total must be
\[
4 \times 95 = 380.
\]

Her first two scores are 97 and 91, which sum to
\[
97 + 91 = 188.
\]

Let her third score be \(x\), and her fourth score be at most 100. To still be able to reach the goal, the total of the last two tests must be at least
\[
380 - 188 = 192.
\]

So if the third test score is \(x\), then the fourth test would need to be at least
\[
192 - x.
\]
Since the maximum possible score on the fourth test is 100, we need
\[
192 - x \le 100.
\]
Thus
\[
x \ge 92.
\]

The lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the shorter tree be \(3x\) and the taller tree be \(4x\), since their heights are in the ratio \(3:4\).

The difference is 16 feet:
\[
4x - 3x = x = 16.
\]

So:
- shorter tree \(= 3(16) = 48\)
- taller tree \(= 4(16) = 64\)

Therefore the taller tree is 64 feet tall.

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.

Starting with 16 teams, to get one winner, we must eliminate 15 teams.

So the number of games played is:
\[
16 - 1 = 15.
\]

ANSWER 8: D

Problem 9:
We need positive integers \(a<b<c\) with
\[
abc = 100.
\]

Factor 100:
\[
100 = 2^2 \cdot 5^2.
\]

We need three positive integer factors in strictly increasing order. Try possible factorizations:

- \(1, 2, 50\) works since \(1<2<50\)
- \(1, 4, 25\) works since \(1<4<25\)
- \(1, 5, 20\) works since \(1<5<20\)
- \(1, 10, 10\) does not work because \(b<c\) fails
- \(2, 2, 25\) does not work because \(a<b\) fails
- \(2, 5, 10\) works since \(2<5<10\)

So there are 4 ways.

ANSWER 9: E

Problem 10:
We want the number of three-digit numbers divisible by 13.

The smallest three-digit number is 100, and the largest is 999.

Find the first multiple of 13 at least 100:
\[
13 \cdot 8 = 104
\]

Find the last multiple of 13 at most 999:
\[
13 \cdot 76 = 988
\]
since
\[
13 \cdot 77 = 1001
\]
is too large.

So the multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \ldots,\ 13\cdot 76
\]

Count them:
\[
76 - 8 + 1 = 69.
\]

ANSWER 10: C

Problem 11:
We want the largest possible difference between two numbers from
\[
\{-16,-4,0,2,4,12\}.
\]

To maximize a difference, subtract the smallest number from the largest number:
\[
12 - (-16) = 28.
\]

So the largest difference is 28.

ANSWER 11: D

Problem 12:
When he was a boy, he ran 15 miles in 3 hours 30 minutes.

Convert 3 hours 30 minutes to hours:
\[
3.5 \text{ hours}
\]

Time per mile then:
\[
\frac{3.5}{15} \text{ hours per mile} = \frac{7}{30} \text{ hours per mile}
\]
Convert to minutes:
\[
\frac{7}{30}\times 60 = 14 \text{ minutes per mile}
\]

Now he walks 10 miles in 4 hours, so time per mile now:
\[
\frac{4}{10} = 0.4 \text{ hours per mile}
\]
\[
0.4 \times 60 = 24 \text{ minutes per mile}
\]

Difference:
\[
24 - 14 = 10
\]

So it takes him 10 minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We must order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
A good way is to compare each pair by cross-multiplying.

Compare \( \frac{15}{11}\) and \( \frac{17}{13}\):  
\[
\frac{15}{11} < \frac{17}{13}
\iff 15\cdot 13 < 17\cdot 11
\]
\[
195 < 187 \quad \text{(false)}
\]
So,
\[
\frac{15}{11} > \frac{17}{13}.
\]

Compare \( \frac{17}{13}\) and \( \frac{19}{15}\):  
\[
\frac{17}{13} < \frac{19}{15}
\iff 17\cdot 15 < 19\cdot 13
\]
\[
255 < 247 \quad \text{(false)}
\]
So,
\[
\frac{17}{13} > \frac{19}{15}.
\]

Now we know:
\[
\frac{19}{15} < \frac{17}{13} < \frac{15}{11}.
\]
This matches choice **E**.

ANSWER 1: E

---

Problem 2:  
There are \(39\) students total. Each has at least one of: dog, cat, or both.  
Let:
- \(D\) = students with a dog
- \(C\) = students with a cat

Given:
\[
|D|=20,\quad |C|=26,\quad |D\cup C|=39.
\]
Use inclusion-exclusion:
\[
|D\cup C|=|D|+|C|-|D\cap C|.
\]
So,
\[
39=20+26-|D\cap C|
\]
\[
39=46-|D\cap C|
\]
\[
|D\cap C|=46-39=7.
\]

ANSWER 2: A

---

Problem 3:  
Start with \(\$100\).  
First year: 15% loss means multiply by \(0.85\).
\[
100(0.85)=85.
\]
Second year: 20% gain on the remaining amount means multiply by \(1.20\).
\[
85(1.20)=102.
\]
Change from original \(\$100\) to \(\$102\) is:
\[
102-100=2 \quad \Rightarrow \quad 2\% \text{ gain}.
\]

ANSWER 3: D

---

Problem 4:  
Wind chill formula:
\[
\text{wind chill} = T - 0.7v
\]
Given \(T=36^\circ\text{F}\), \(v=18\) mph:
\[
36 - 0.7(18)=36 - 12.6=23.4.
\]
Closest choice is \(23\).

ANSWER 4: B

---

Problem 5:  
“All but 6 are red” means number of red marbles is \(6\).  
Similarly:
- All but 6 are red \(\Rightarrow R=6\)
- All but 8 are green \(\Rightarrow G=8\)
- All but 4 are blue \(\Rightarrow B=4\)

But the jar contains marbles that are each exactly one color, so the total is:
\[
R+G+B = 6+8+4 = 18.
\]
However, notice the statement “all but 6 are red marbles” means:  
Among all marbles, \( \text{(total)} - 6\) are red. So red is \(\text{total}-6\), not 6.

Let total be \(N\). Then:
- Red: \(N-6\)
- Green: \(N-8\)
- Blue: \(N-4\)

Since all marbles are exactly one color,
\[
(N-6)+(N-8)+(N-4)=N.
\]
\[
3N - (6+8+4)=N
\]
\[
3N-18=N
\]
\[
2N=18
\]
\[
N=9.
\]

ANSWER 5: C

---

Problem 6:  
Goal average \(95\) over 4 tests \(\Rightarrow\) total needed:
\[
4\cdot 95=380.
\]
First two scores: \(97\) and \(91\), sum:
\[
97+91=188.
\]
After two tests, remaining needed for last two tests is:
\[
380-188=192.
\]
Let the third test score be \(x\). The fourth test can be at most \(100\).  
To make \(x\) as low as possible, make the fourth as high as possible: \(100\).

Then:
\[
x+100=192 \Rightarrow x=92.
\]

ANSWER 6: B

---

Problem 7:  
Let tree heights be in ratio \(3:4\).  
So taller tree is \(4k\) and shorter tree is \(3k\).

Top of one tree is 16 ft higher than the top of the other:
\[
4k-3k=16 \Rightarrow k=16.
\]
Taller tree height:
\[
4k=4\cdot 16=64.
\]

ANSWER 7: B

---

Problem 8:  
Tournament with 16 teams, losing team eliminated each game (single-elimination).  
Number of games played to determine a winner is always:
\[
\text{teams} - 1 = 16-1 = 15.
\]

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a<b<c\) such that:
\[
abc=100.
\]
Factor:
\[
100=2^2\cdot 5^2.
\]
We want all factorizations into three positive integers with strict inequality.

List triples \((a,b,c)\) with \(a<b<c\):

First find all ways to write 100 as \(a\cdot b\cdot c\) with \(a\) smallest.

Try \(a=1\): then \(bc=100\) with \(b<c\). Factor pairs of 100:
\[
100=2\cdot 50,\ 4\cdot 25,\ 5\cdot 20,\ 10\cdot 10.
\]
Need \(b<c\), so ignore \(10\cdot 10\). We get:
\[
(1,2,50),\ (1,4,25),\ (1,5,20)
\]
\((1,10,10)\) not allowed.

Now \(a=2\): then \(bc=50\). Factor pairs:
\[
50=2\cdot 25,\ 5\cdot 10.
\]
Need \(b<c\) and also \(a<b\) so \(2<b\).
- \(b=2\) would fail since \(a=b\), so use \(5\cdot 10\):
\[
(2,5,10)
\]

Now \(a=4\): then \(bc=25\). Factors:
\[
25=5\cdot 5
\]
But \(b=c\), not allowed.

\(a=5\): then \(bc=20\) and \(b>5\) would force \(b=10,c=2\) not possible. No more.

Total valid sets:  
\[
(1,2,50),\ (1,4,25),\ (1,5,20),\ (2,5,10) \Rightarrow 4 \text{ ways}.
\]

ANSWER 9: E

---

Problem 10:  
Count three-digit numbers divisible by 13.

Three-digit range: 100 to 999.  
Find the smallest multiple of 13 ≥ 100 and largest ≤ 999.

\(13\cdot 8=104\) (smallest since \(13\cdot 7=91\)).  
\(13\cdot 76=988\) and \(13\cdot 77=1001\) too big, so largest is \(988=13\cdot 76\).

So multiples are:
\[
13\cdot 8,\ 13\cdot 9,\ \dots,\ 13\cdot 76
\]
Count:
\[
76-8+1=69.
\]

ANSWER 10: C

---

Problem 11:  
Set: \(\{-16,-4,0,2,4,12\}\).  
The largest difference from subtracting two chosen numbers is achieved by
\[
\text{(largest)} - \text{(smallest)} = 12 - (-16)=28.
\]

ANSWER 11: D

---

Problem 12:  
When boy: run 15 miles in 3 hours 30 minutes.  
\(3\) hours \(30\) minutes \(= 3.5\) hours.  
Time per mile then:
\[
\frac{3.5}{15}\text{ hours per mile}=\frac{7/2}{15}=\frac{7}{30}\text{ hours}.
\]
Convert to minutes:
\[
\frac{7}{30}\cdot 60=14\text{ minutes per mile}.
\]

As old man: walk 10 miles in 4 hours.  
Time per mile:
\[
\frac{4}{10}=0.4\text{ hours per mile}.
\]
Convert to minutes:
\[
0.4\cdot 60=24\text{ minutes per mile}.
\]

Difference:
\[
24-14=10\text{ minutes}.
\]

ANSWER 12: B

Problem 1:
To compare $\frac{15}{11}, \frac{19}{15},$ and $\frac{17}{13}$, we can write them as $1 + \frac{4}{11}, 1 + \frac{4}{15},$ and $1 + \frac{4}{13}$. Since the numerators are identical, the fraction with the largest denominator is the smallest. Comparing the denominators $11 < 13 < 15$, we find that $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Thus, $1 + \frac{4}{15} < 1 + \frac{4}{13} < 1 + \frac{4}{11}$, which is $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
ANSWER 1: E

Problem 2:
Using the Principle of Inclusion-Exclusion, let $D$ be the set of students with a dog and $C$ be the set of students with a cat. We have $|D| = 20$, $|C| = 26$, and $|D \cup C| = 39$. The number of students with both is $|D \cap C| = |D| + |C| - |D \cup C| = 20 + 26 - 39 = 46 - 39 = 7$.
ANSWER 2: A

Problem 3:
Tammy starts with $100. After a 15% loss, she has $100 \times (1 - 0.15) = 100 \times 0.85 = 85$. After a 20% gain on the remaining $85, she has $85 \times (1 + 0.20) = 85 \times 1.2 = 102$. The change from $100 to $102 is a $2 gain, which is a 2% gain.
ANSWER 3: D

Problem 4:
The formula is $36 - 0.7 \times 18$. First, $0.7 \times 18 = 7 \times 1.8 = 12.6$. Then, $36 - 12.6 = 23.4$. The value closest to 23.4 among the choices is 23.
ANSWER 4: B

Problem 5:
Let $R, G, B$ be the number of red, green, and blue marbles, and $T$ be the total. $T-R=6, T-G=8, T-B=4$. Adding these: $3T - (R+G+B) = 18$. Since $R+G+B=T$, we have $3T - T = 18$, so $2T = 18$, $T = 9$.
ANSWER 5: C

Problem 6:
Isabella needs a total of $95 \times 4 = 380$ points. Her first two tests total $97 + 91 = 188$. She needs $380 - 188 = 192$ points from the last two tests. To find the lowest possible score on the third test, we assume she gets the maximum score (100) on the fourth test. Thus, the third test must be at least $192 - 100 = 92$.
ANSWER 6: B

Problem 7:
Let the heights be $3x$ and $4x$. The difference is $4x - 3x = x$. We are given $x = 16$. The taller tree is $4x = 4 \times 16 = 64$.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game results in exactly one team being eliminated. To have one winner among 16 teams, 15 teams must be eliminated. Therefore, 15 games must be played.
ANSWER 8: D

Problem 9:
We need $a \times b \times c = 100$ with $a < b < c$. Possible sets $\{a, b, c\}$:
1. $\{1, 2, 50\}$
2. $\{1, 4, 25\}$
3. $\{1, 5, 20\}$
4. $\{1, 10, 10\}$ (No, $b$ must be $< c$)
5. $\{2, 5, 10\}$
There are 4 such sets.
ANSWER 9: E

Problem 10:
The smallest three-digit number divisible by 13 is $13 \times 8 = 104$. The largest is $13 \times 76 = 988$. The number of multiples is $76 - 8 + 1 = 69$.
ANSWER 10: C

Problem 11:
The largest difference is the maximum value minus the minimum value. Max = 12, Min = -16. Difference = $12 - (-16) = 12 + 16 = 28$.
ANSWER 11: D

Problem 12:
Boy: 15 miles in 3.5 hours (210 minutes). Time per mile = $210 / 15 = 14$ minutes.
Man: 10 miles in 4 hours (240 minutes). Time per mile = $240 / 10 = 24$ minutes.
Difference = $24 - 14 = 10$ minutes.
ANSWER 12: B

Problem 1:
To order the fractions \(15/11\), \(19/15\), and \(17/13\) from least to greatest, convert each to a decimal approximation by division:
\(15/11 \approx 1.3636\),
\(19/15 \approx 1.2667\),
\(17/13 \approx 1.3077\).
The resulting order is \(19/15 < 17/13 < 15/11\), which matches choice E. (Cross-multiplication yields the same inequalities: \(19 \times 13 = 247 < 15 \times 17 = 255\), \(17 \times 15 = 255 < 13 \times 19 = 247\) is false so \(17/13 > 19/15\), and \(15 \times 15 = 225 < 11 \times 17 = 187\) is false so \(15/11 > 17/13\).)
ANSWER 1: E
Problem 2:
Let \(B\) be the number of students with both pets. The total number of students owning at least one pet satisfies the inclusion-exclusion equation
\(20 + 26 - B = 39\).
Simplifying gives \(46 - B = 39\), so \(B = 7\). This matches choice A.
ANSWER 2: A
Problem 3:
After a 15% loss the investment is \(100 \times 0.85 = 85\). After a subsequent 20% gain it becomes \(85 \times 1.2 = 102\). The net change from the original 100 is therefore a gain of 2, which matches choice D.
ANSWER 3: D
Problem 4:
Substitute the given values into the wind-chill formula:
\(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\).
The value 23.4 is closest to 23, which matches choice B.
ANSWER 4: B
Problem 5:
Let \(T\) be the total number of marbles. Then the number of red marbles is \(T-6\), green is \(T-8\), and blue is \(T-4\). Because every marble has exactly one color,
\((T-6) + (T-8) + (T-4) = T\).
Simplifying yields \(3T - 18 = T\), so \(2T = 18\) and \(T = 9\). Direct verification confirms the counts satisfy the original statements, matching choice C.
ANSWER 5: C
Problem 6:
A 95 average on four 100-point tests requires a total of 380 points. The first two scores sum to 188, so the remaining two scores must sum to 192. The lowest possible third-test score that still permits a fourth-test score of at most 100 is therefore \(192 - 100 = 92\), which matches choice B.
ANSWER 6: B
Problem 7:
The heights differ by 16 ft and are in the ratio 3 : 4. The difference corresponds to one part of the ratio, so each part equals 16 ft. The taller tree is therefore \(4 \times 16 = 64\) ft, which matches choice B.
ANSWER 7: B
Problem 8:
A single-elimination tournament with 16 teams requires the elimination of 15 teams to produce a single champion. Each game eliminates exactly one team, so exactly 15 games are needed, which matches choice D.
ANSWER 8: D
Problem 9:
The positive-integer triples \((a,b,c)\) with \(a < b < c\) and \(a \cdot b \cdot c = 100\) are
\((1,4,25)\), \((1,5,20)\), \((1,2,50)\), and \((2,5,10)\).
There are exactly four such ordered triples, which matches choice E.
ANSWER 9: E
Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\) and the largest is \(13 \times 76 = 988\). The integers from 8 to 76 inclusive number \(76 - 8 + 1 = 69\), which matches choice C.
ANSWER 10: C
Problem 11:
The largest possible difference is obtained by subtracting the smallest element from the largest element:
\(12 - (-16) = 28\),
which matches choice D.
ANSWER 11: D
Problem 12:
As a boy, the time per mile is \(210\) minutes \(\div 15 = 14\) minutes. As an old man, the time per mile is \(240\) minutes \(\div 10 = 24\) minutes. The difference is \(24 - 14 = 10\) minutes, which matches choice B.
ANSWER 12: B

## Step 1: Problem 1 - Restate the problem
We need to determine the correct order of the fractions 15/11, 19/15, and 17/13 from least to greatest.

## Step 2: Problem 1 - Compare the fractions
To compare these fractions, we can convert them to equivalent decimals or find a common denominator. Let's convert them to decimals: 15/11 = 1.3636..., 19/15 = 1.2666..., and 17/13 = 1.3076....

## Step 3: Problem 1 - Order the fractions
Now, we compare the decimals: 1.2666... < 1.3076... < 1.3636..., so 19/15 < 17/13 < 15/11.

## Step 4: Problem 1 - Match the order to the given choices
The correct order from least to greatest is 19/15 < 17/13 < 15/11, which corresponds to choice E.

## Step 5: Problem 1 - Provide the answer
ANSWER 1: E

## Step 6: Problem 2 - Restate the problem
We are given that 39 students have either a dog, a cat, or both, with 20 having a dog and 26 having a cat. We need to find how many students have both.

## Step 7: Problem 2 - Apply the principle of inclusion-exclusion
The total number of students with a dog or a cat is given by (number of students with a dog) + (number of students with a cat) - (number of students with both), because those with both are counted twice. So, 20 + 26 - (number with both) = 39.

## Step 8: Problem 2 - Calculate the number of students with both a dog and a cat
46 - (number with both) = 39. Therefore, the number with both = 46 - 39 = 7.

## Step 9: Problem 2 - Provide the answer
ANSWER 2: A

## Step 10: Problem 3 - Restate the problem
Tycoon Tammy invested $100. It decreased by 15% in the first year and then increased by 20% in the second year. We need to calculate the overall change.

## Step 11: Problem 3 - Calculate the value after the first year
After a 15% loss, the investment is 100 * (1 - 0.15) = 100 * 0.85 = $85.

## Step 12: Problem 3 - Calculate the value after the second year
After a 20% gain, the investment is 85 * (1 + 0.20) = 85 * 1.20 = $102.

## Step 13: Problem 3 - Determine the overall change
The overall change is $102 - $100 = $2 gain, which is a 2% gain.

## Step 14: Problem 3 - Provide the answer
ANSWER 3: D

## Step 15: Problem 4 - Restate the problem
Given the air temperature is 36°F and the wind speed is 18 mph, we need to calculate the wind chill using the formula: wind chill = air temperature - 0.7 * wind speed.

## Step 16: Problem 4 - Calculate the wind chill
Wind chill = 36 - 0.7 * 18 = 36 - 12.6 = 23.4°F.

## Step 17: Problem 4 - Round to the nearest answer choice
The closest value to 23.4 is 23.

## Step 18: Problem 4 - Provide the answer
ANSWER 4: B

## Step 19: Problem 5 - Restate the problem
There are red, green, and blue marbles in a jar. All but 6 are red, all but 8 are green, and all but 4 are blue. We need to find the total number of marbles.

## Step 20: Problem 5 - Set up equations based on the given information
Let's denote the total number of marbles as T, red marbles as R, green marbles as G, and blue marbles as B. We have: T - R = 6, T - G = 8, and T - B = 4.

## Step 21: Problem 5 - Sum the equations
(T - R) + (T - G) + (T - B) = 6 + 8 + 4, which simplifies to 3T - (R + G + B) = 18. Since R + G + B = T, we have 3T - T = 18.

## Step 22: Problem 5 - Solve for T
2T = 18, so T = 9.

## Step 23: Problem 5 - Provide the answer
ANSWER 5: C

## Step 24: Problem 6 - Restate the problem
Isabella needs an average of 95 on four 100-point tests. Her first two scores were 97 and 91. We need to find the lowest possible score on the third test that still allows her to reach her goal.

## Step 25: Problem 6 - Calculate the total score needed for an average of 95
The total score needed for an average of 95 on four tests is 95 * 4 = 380.

## Step 26: Problem 6 - Calculate the score needed on the last two tests
After the first two tests, Isabella has 97 + 91 = 188 points. So, she needs 380 - 188 = 192 points from the last two tests.

## Step 27: Problem 6 - Determine the lowest possible score on the third test
To minimize the third test score, we maximize the fourth test score. The maximum score on a test is 100. So, the score on the fourth test is 100, and the score on the third test is 192 - 100 = 92.

## Step 28: Problem 6 - Provide the answer
ANSWER 6: B

## Step 29: Problem 7 - Restate the problem
The tops of two trees are 16 feet apart in height, and their heights are in the ratio 3:4. We need to find the height of the taller tree.

## Step 30: Problem 7 - Set up the ratio and the height difference
Let's denote the heights of the trees as 3x and 4x. The difference in their heights is 4x - 3x = x = 16.

## Step 31: Problem 7 - Calculate the height of the taller tree
Since x = 16, the taller tree's height is 4x = 4 * 16 = 64 feet.

## Step 32: Problem 7 - Provide the answer
ANSWER 7: B

## Step 33: Problem 8 - Restate the problem
There are 16 teams in a tournament where the losing team is eliminated. We need to find out how many games will be played to determine the winner.

## Step 34: Problem 8 - Understand the tournament structure
In a single-elimination tournament, each game eliminates one team. All teams but one (the winner) must be eliminated.

## Step 35: Problem 8 - Calculate the number of games needed
To have one winner out of 16 teams, 15 teams must be eliminated, which requires 15 games.

## Step 36: Problem 8 - Provide the answer
ANSWER 8: D

## Step 37: Problem 9 - Restate the problem
We need to find the number of ways to choose three positive integers a, b, and c such that a < b < c and their product is 100.

## Step 38: Problem 9 - Factorize 100
100 = 2^2 * 5^2. We need to distribute these factors among a, b, and c.

## Step 39: Problem 9 - List possible combinations
Possible combinations where a < b < c include: (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10) - which is not valid because b = c, (2, 5, 10), and (4, 5, 5) - which is also not valid.

## Step 40: Problem 9 - Count the valid combinations
Valid combinations are (1, 2, 50), (1, 4, 25), (1, 5, 20), and (2, 5, 10), making 4 valid combinations

Problem 1:
We need to order the fractions \(\frac{15}{11}\), \(\frac{19}{15}\), and \(\frac{17}{13}\) from least to greatest. Convert each to a decimal:
\(\frac{15}{11} \approx 1.3636\)
\(\frac{19}{15} \approx 1.2667\)
\(\frac{17}{13} \approx 1.3077\)
So the order is \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\), which matches choice E.
ANSWER 1: E

Problem 2:
Use the inclusion-exclusion principle: Total students = (students with a dog) + (students with a cat) – (students with both).
\(39 = 20 + 26 - \text{both}\) → \(\text{both} = 46 - 39 = 7\).
ANSWER 2: A

Problem 3:
Start with $100. After a 15% loss: \(100 \times 0.85 = 85\).
After a 20% gain on the remaining: \(85 \times 1.20 = 102\).
The change is \(+2\), which is a 2% gain.
ANSWER 3: D

Problem 4:
Wind chill = air temperature – \(0.7 \times\) wind speed.
Wind chill = \(36 - 0.7 \times 18 = 36 - 12.6 = 23.4\), which is closest to 23.
ANSWER 4: B

Problem 5:
Let \(R, G, B\) be the number of red, green, and blue marbles.
"All but 6 are red" means \(G + B = 6\).
"All but 8 are green" means \(R + B = 8\).
"All but 4 are blue" means \(R + G = 4\).
Add the three equations: \(2(R+G+B) = 18\) → \(R+G+B = 9\).
Total marbles = 9.
ANSWER 5: C

Problem 6:
Target total points = \(4 \times 95 = 380\).
Points from first two tests = \(97 + 91 = 188\).
Points needed from the last two tests = \(380 - 188 = 192\).
To still have a chance after the third test, even a perfect 100 on the fourth test must be enough. So the third test score must be at least \(192 - 100 = 92\).
ANSWER 6: B

Problem 7:
Let the heights be \(3x\) and \(4x\). Their difference is \(x = 16\).
The taller tree is \(4x = 64\) feet.
ANSWER 7: B

Problem 8:
In a single-elimination tournament, each game eliminates one team. To go from 16 teams to 1 winner, 15 teams must be eliminated, so 15 games are played.
ANSWER 8: D

Problem 9:
We need triples \((a,b,c)\) of positive integers with \(a < b < c\) and \(abc = 100\).
Factor 100 = \(2^2 \cdot 5^2\). List all sets of three distinct factors whose product is 100:
\((1, 2, 50)\), \((1, 4, 25)\), \((1, 5, 20)\), \((2, 5, 10)\).
That gives 4 ways.
ANSWER 9: E

Problem 10:
The smallest three-digit multiple of 13 is \(13 \times 8 = 104\).
The largest three-digit multiple of 13 is \(13 \times 76 = 988\).
Number of multiples = \(76 - 8 + 1 = 69\).
ANSWER 10: C

Problem 11:
To maximize the difference \(a - b\), choose the largest number for \(a\) and the smallest for \(b\).
From the set \(\{-16, -4, 0, 2, 4, 12\}\), max = 12, min = –16.
Difference = \(12 - (-16) = 28\).
ANSWER 11: D

Problem 12:
As a boy: 15 miles in 3 hours 30 minutes = 210 minutes → \(210 \div 15 = 14\) minutes per mile.
As an old man: 10 miles in 4 hours = 240 minutes → \(240 \div 10 = 24\) minutes per mile.
Difference = \(24 - 14 = 10\) minutes longer per mile.
ANSWER 12: B

Problem 1:
We need to order the fractions 15/11, 19/15, and 17/13 from least to greatest.

Each fraction is greater than 1, so let's write them as mixed numbers to compare easily:
- 15/11 = 1 + 4/11 ≈ 1.3636...
- 19/15 = 1 + 4/15 ≈ 1.2667...
- 17/13 = 1 + 4/13 ≈ 1.3077...

Notice that each fraction has the form (n+4)/n = 1 + 4/n. Since the numerator of the fractional part is always 4, the larger the denominator n, the smaller the fraction. The denominators are 15, 13, and 11.

Since 15 > 13 > 11, we get 4/15 < 4/13 < 4/11, so:
19/15 < 17/13 < 15/11

This matches choice E.

ANSWER 1: E

Problem 2:
We use the inclusion-exclusion principle. Let D = students with a dog, C = students with a cat, and B = students with both.

Every student has at least one pet, so:
Total = D + C − B
39 = 20 + 26 − B
39 = 46 − B
B = 46 − 39 = 7

Choice A is 7. (Choice E, 46, is a trap — it's just 20+26 without subtracting the overlap.)

ANSWER 2: A

Problem 3:
Start with $100.

Year 1: 15% loss → multiply by (1 − 0.15) = 0.85
$100 × 0.85 = $85

Year 2: 20% gain on the remaining $85 → multiply by (1 + 0.20) = 1.20
$85 × 1.20 = $102

Overall change: $102 − $100 = +$2, which is a 2% gain on the original $100.

(A common trap is to add −15% + 20% = +5%, but percentages apply to different bases.)

ANSWER 3: D

Problem 4:
Using the formula: Wind chill = (air temperature) − 0.7 × (wind speed)

Plug in the values:
Wind chill = 36 − 0.7 × 18
= 36 − 12.6
= 23.4

The closest answer choice is 23.

ANSWER 4: B

Problem 5:
Let R, G, B be the number of red, green, and blue marbles, and T = R + G + B be the total.

- "All but 6 are red" → G + B = 6 → T − R = 6
- "All but 8 are green" → R + B = 8 → T − G = 8
- "All but 4 are blue" → R + G = 4 → T − B = 4

Add all three equations:
(G + B) + (R + B) + (R + G) = 6 + 8 + 4
2R + 2G + 2B = 18
2T = 18
T = 9

(Check: R = 3, G = 1, B = 5. All but 6 are red: 1+5=6 ✓. All but 8 are green: 3+5=8 ✓. All but 4 are blue: 3+1=4 ✓.)

ANSWER 5: C

Problem 6:
Isabella needs an average of 95 on 4 tests, so the total points needed = 95 × 4 = 380.

Her first two scores: 97 + 91 = 188.
Let x = third test score, y = fourth test score.
188 + x + y = 380
x + y = 192

To minimize x, we maximize y. The maximum possible test score is 100.
x + 100 = 192
x = 92

So the lowest possible third test score is 92.

ANSWER 6: B

Problem 7:
Let the heights of the two trees be 3k and 4k (ratio 3:4).

The taller tree is 16 feet higher:
4k − 3k = 16
k = 16

The taller tree = 4k = 4 × 16 = 64 feet.

(A trap is choosing 48, which is the shorter tree 3×16.)

ANSWER 7: B

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team. To go from 16 teams down to 1 winner, we must eliminate 15 teams.

Therefore, exactly 15 games must be played.

(We can verify: Round 1: 8 games → 8 teams. Round 2: 4 games → 4 teams. Round 3: 2 games → 2 teams. Final: 1 game → 1 winner. Total = 8+4+2+1 = 15.)

ANSWER 8: D

Problem 9:
We need positive integers a < b < c such that a × b × c = 100.

First, 100 = 2² × 5². The divisors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Systematically check values of a:

**a = 1:** b × c = 100, with 1 < b < c
- b = 2, c = 50 → (1, 2, 50) ✓
- b = 4, c = 25 → (1, 4, 25) ✓
- b = 5, c = 20 → (1, 5, 20) ✓
- b = 10, c = 10 → not valid (b must be < c)

**a = 2:** b × c = 50, with 2 < b < c
- b = 5, c = 10 → (2, 5, 10) ✓

**a = 4:** b × c = 25, with 4 < b < c
- b = 5, c = 5 → not valid (b < c fails)

**a = 5:** b × c = 20, with 5 < b < c → no integer solutions.

Total: 4 ways.

ANSWER 9: E

Problem 10:
Three-digit numbers range from 100 to 999.

Smallest three-digit multiple of 13: ⌈100/13⌉ = ⌈7.69⌉ = 8, so 13 × 8 = 104.
Largest three-digit multiple of 13: ⌊999/13⌋ = ⌊76.84⌋ = 76, so 13 × 76 = 988.

Number of multiples = 76 − 8 + 1 = 69.

ANSWER 10: C

Problem 11:
Set: {−16, −4, 0, 2, 4, 12}

To maximize the difference (first number − second number), we want the largest possible first number and the smallest possible second number.

Largest number in the set: 12
Smallest number in the set: −16

Largest difference = 12 − (−16) = 12 + 16 = 28.

(A tempting wrong answer is 16, from 12 − (−4), but we must use −16.)

ANSWER 11: D

Problem 12:
**As a boy:** 15 miles in 3 hours 30 minutes = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes/mile.

**As an old man:** 10 miles in 4 hours = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes/mile.

**Difference:** 24 − 14 = 10 minutes longer per mile now.

ANSWER 12: B

 Problem 1:
We are asked to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, and $\frac{17}{13}$ from least to greatest.  
Each fraction is slightly larger than $1$, so rewrite each as $1$ plus a proper fraction:
$$\frac{15}{11}=1+\frac{4}{11},\qquad \frac{19}{15}=1+\frac{4}{15},\qquad \frac{17}{13}=1+\frac{4}{13}.$$
Because the integer parts are equal, we compare $\frac{4}{11}$, $\frac{4}{15}$, and $\frac{4}{13}$.  
For positive fractions with the same numerator, the one with the **larger** denominator is **smaller**. Thus
$$\frac{4}{15}<\frac{4}{13}<\frac{4}{11},$$
which means
$$\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.$$
Checking the choices:
- A and B incorrectly place $\frac{15}{11}$ as the smallest.
- C reverses $\frac{19}{15}$ and $\frac{17}{13}$.
- D places $\frac{15}{11}$ in the middle, but it is the largest.

ANSWER 1: E

---

Problem 2:
We have $39$ students. Each student has a dog, a cat, or both. $20$ have a dog and $26$ have a cat. We need the number who have both.  
Use the principle of inclusion–exclusion:
$$|D\cup C|=|D|+|C|-|D\cap C|.$$
Since every student has at least one pet, $|D\cup C|=39$. Substituting:
$$39=20+26-|D\cap C|$$
$$39=46-|D\cap C|$$
$$|D\cap C|=7.$$
Thus $7$ students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:
Tammy invests $\$100$. In year one she loses $15\%$; in year two she gains $20\%$ on the remaining amount. We need the overall percent change.  
After year 1:
$$100\times(1-0.15)=100\times0.85=\$85.$$
After year 2:
$$85\times(1+0.20)=85\times1.20=\$102.$$
The final value is $\$102$, a gain of $\$2$ on the original $\$100$, which is a $2\%$ gain.  
(Note: a common wrong choice is $5\%$ gain, which incorrectly adds the percentages without accounting for the reduced principal after the loss.)

ANSWER 3: D

---

Problem 4:
The wind chill formula is
$$\text{wind chill}=\text{air temperature}-0.7\times(\text{wind speed}).$$
With air temperature $36^\circ\text{F}$ and wind speed $18$ mph:
$$\text{wind chill}=36-0.7(18)=36-12.6=23.4.$$
The value closest to $23.4$ is $23$.

ANSWER 4: B

---

Problem 5:
Let the numbers of red, green, and blue marbles be $R$, $G$, $B$, and let the total be $T=R+G+B$.  
"All but $6$ are red" means the non-red marbles total $6$: $G+B=6$.  
"All but $8$ are green" means $R+B=8$.  
"All but $4$ are blue" means $R+G=4$.  
Add these three equations:
$$(G+B)+(R+B)+(R+G)=6+8+4$$
$$2R+2G+2B=18$$
$$2T=18\implies T=9.$$
(Checking: $R=3$, $G=1$, $B=5$, and $3+1+5=9$.)

ANSWER 5: C

---

Problem 6:
Isabella wants an average of $95$ on four $100$-point tests. The required total is
$$4\times95=380.$$
Her first two scores sum to $97+91=188$.  
She needs $380-188=192$ points from the third and fourth tests combined.  
To find the **lowest** possible third-test score, assume she earns the **maximum** $100$ on the fourth test:
$$\text{third test}=192-100=92.$$
If she scored $91$ on the third test, she would need $101$ on the fourth, which is impossible. Thus $92$ is the minimum.

ANSWER 6: B

---

Problem 7:
Let the tree heights be $3x$ and $4x$ (ratio $3:4$).  
The difference in height is
$$4x-3x=x=16\text{ feet}.$$
Therefore the taller tree is
$$4x=4(16)=64\text{ feet}.$$

ANSWER 7: B

---

Problem 8:
In a single-elimination tournament, every game eliminates exactly one team.  
Starting with $16$ teams, we must eliminate $15$ teams to declare $1$ winner.  
Hence $15$ games are required.  
(Alternatively: $8$ games in round 1, $4$ in round 2, $2$ in round 3, $1$ in the final; $8+4+2+1=15$.)

ANSWER 8: D

---

Problem 9:
We need positive integers $a<b<c$ with $abc=100$. Factor $100=2^2\cdot5^2$.  
Systematically list factor triples with distinct factors in increasing order:
- $a=1$: then $bc=100$. Pairs with $b<c$: $(2,50)$, $(4,25)$, $(5,20)$.  
  This gives $(1,2,50)$, $(1,4,25)$, $(1,5,20)$.
- $a=2$: then $bc=50$. Pairs with $2<b<c$: $(5,10)$.  
  This gives $(2,5,10)$.
- $a=4$: then $bc=25$. The only pair is $(5,5)$, but $b<c$ fails.
- Larger $a$ makes $a^3>100$, impossible.

There are exactly $4$ valid triples.

ANSWER 9: E

---

Problem 10:
We want the count of three-digit multiples of $13$.  
The smallest three-digit multiple is $13\times8=104$.  
The largest three-digit multiple is $13\times76=988$ (since $13\times77=1001$).  
The number of multiples is
$$76-8+1=69.$$

ANSWER 10: C

---

Problem 11:
To maximize the difference $a-b$ where $a$ and $b$ are chosen from $\{-16,-4,0,2,4,12\}$, choose the **largest** number for $a$ and the **smallest** for $b$:
$$a=12,\qquad b=-16.$$
$$12-(-16)=12+16=28.$$
(Other combinations such as $12-0=12$ or $4-(-16)=20$ are smaller.)

ANSWER 11: D

---

Problem 12:
As a boy: $15$ miles in $3$ hours $30$ minutes $=210$ minutes.  
Time per mile: $\frac{210}{15}=14$ minutes.  
As an old man: $10$ miles in $4$ hours $=240$ minutes.  
Time per mile: $\frac{240}{10}=24$ minutes.  
The difference is $24-14=10$ minutes longer per mile now.

ANSWER 12: B

Problem 1:  
We need to order the fractions \( \frac{15}{11}, \frac{19}{15}, \frac{17}{13} \) from least to greatest.  
Compare using cross-multiplication:  

- Compare \(\frac{15}{11}\) and \(\frac{17}{13}\): \(15 \times 13 = 195\), \(11 \times 17 = 187\). Since \(195 > 187\), \(\frac{15}{11} > \frac{17}{13}\).  
- Compare \(\frac{17}{13}\) and \(\frac{19}{15}\): \(17 \times 15 = 255\), \(13 \times 19 = 247\). Since \(255 > 247\), \(\frac{17}{13} > \frac{19}{15}\).  

Thus \(\frac{19}{15} < \frac{17}{13} < \frac{15}{11}\).  
This matches choice E.

ANSWER 1: E

---

Problem 2:  
Let \(D\) be the set of students with a dog, \(C\) with a cat.  
Given: \(|D| = 20\), \(|C| = 26\), total students = 39, and every student has at least one pet.  
By inclusion–exclusion: \(|D \cup C| = |D| + |C| - |D \cap C|\).  
So \(39 = 20 + 26 - |D \cap C|\) → \(|D \cap C| = 46 - 39 = 7\).  
Thus 7 students have both a dog and a cat.

ANSWER 2: A

---

Problem 3:  
Initial investment: $100.  
After a 15% loss: \(100 \times (1 - 0.15) = 100 \times 0.85 = 85\).  
After a 20% gain on the remaining: \(85 \times (1 + 0.20) = 85 \times 1.2 = 102\).  
Final amount = $102, which is a $2 increase over $100.  
Percent change = \(\frac{2}{100} \times 100\% = 2\%\) gain.  
So the change is a 2% gain.

ANSWER 3: D

---

Problem 4:  
Wind chill formula: \(\text{wind chill} = \text{air temperature} - 0.7 \times \text{wind speed}\).  
Given air temperature = 36°F, wind speed = 18 mph.  
Compute: \(0.7 \times 18 = 12.6\).  
Wind chill = \(36 - 12.6 = 23.4\).  
Closest option is 23.

ANSWER 4: B

---

Problem 5:  
Let total number of marbles be \(T\).  
“All but 6 are red” means non‑red = 6, so red = \(T - 6\).  
Similarly, green = \(T - 8\), blue = \(T - 4\).  
Total = red + green + blue = \((T-6)+(T-8)+(T-4) = 3T - 18\).  
Set equal to \(T\): \(T = 3T - 18\) → \(2T = 18\) → \(T = 9\).  
Thus there are 9 marbles.

ANSWER 5: C

---

Problem 6:  
Isabella wants an average of 95 over four tests.  
Scores: 97, 91, \(x\) (third test), \(y\) (fourth test, maximum 100).  
Average condition: \(\frac{97+91+x+y}{4} = 95\) → \(188 + x + y = 380\) → \(x + y = 192\).  
So \(y = 192 - x\).  
For it to be possible, \(y\) must be between 0 and 100 inclusive: \(0 \le 192 - x \le 100\).  
The right inequality gives \(192 - x \le 100\) → \(x \ge 92\).  
The left inequality gives \(192 - x \ge 0\) → \(x \le 192\) (automatically true since \(x \le 100\)).  
Thus the smallest possible \(x\) is 92 (then \(y = 100\)).  
So the lowest possible third‑test score is 92.

ANSWER 6: B

---

Problem 7:  
Let the heights be \(3k\) and \(4k\) (ratio 3:4).  
The difference is \(4k - 3k = k = 16\) feet.  
Therefore the taller tree is \(4k = 4 \times 16 = 64\) feet.

ANSWER 7: B

---

Problem 8:  
In a single‑elimination tournament, each game eliminates exactly one team.  
To determine a champion from 16 teams, 15 teams must be eliminated.  
Hence the number of games played is 15.  
(Alternatively: 16 → 8 (8 games), 8 → 4 (4 games), 4 → 2 (2 games), 2 → 1 (1 game); total 8+4+2+1 = 15.)

ANSWER 8: D

---

Problem 9:  
We need positive integers \(a, b, c\) with \(a < b < c\) and \(abc = 100\).  
Factor \(100 = 2^2 \cdot 5^2\).  
Enumerate all triples with product 100 and distinct increasing order:  

- \(a = 1\): then \(bc = 100\). Pairs with \(b < c\): \((2,50), (4,25), (5,20)\). → triples \((1,2,50), (1,4,25), (1,5,20)\).  
- \(a = 2\): then \(bc = 50\). Pairs with \(b > 2\) and \(b < c\): \((5,10)\). → triple \((2,5,10)\).  
- \(a = 4\): then \(bc = 25\). No pair with \(b > 4\) and \(b < c\) (only \((5,5)\) but \(b = c\)).  
- \(a = 5\): then \(bc = 20\). No pair with \(b > 5\) and \(b < c\).  
- Larger \(a\) impossible because product would be too small.  

Thus there are 4 ways.

ANSWER 9: E

---

Problem 10:  
Three‑digit numbers are from 100 to 999 inclusive.  
Find smallest multiple of 13 ≥ 100: \(13 \times 7 = 91\), \(13 \times 8 = 104\). So first = 104.  
Largest multiple of 13 ≤ 999: \(13 \times 76 = 988\), \(13 \times 77 = 1001 > 999\). So last = 988.  
Number of multiples = \(76 - 8 + 1 = 69\).

ANSWER 10: C

---

Problem 11:  
To obtain the largest possible result from subtracting one chosen number from another, subtract the smallest number from the largest number.  
Set = \(\{-16, -4, 0, 2, 4, 12\}\).  
Largest = 12, smallest = -16.  
Difference = \(12 - (-16) = 28\).  
No other subtraction gives a larger value.

ANSWER 11: D

---

Problem 12:  
Boy: 15 miles in 3 hours 30 minutes = 3.5 hours = 210 minutes.  
Time per mile = \(210 \div 15 = 14\) minutes.  
Old man: 10 miles in 4 hours = 240 minutes.  
Time per mile = \(240 \div 10 = 24\) minutes.  
Difference = \(24 - 14 = 10\) minutes longer per mile.

ANSWER 12: B

**Problem 1:**  
We compare the three fractions by converting them to decimals (or by cross‑multiplication).

\[
\frac{15}{11}\approx1.3636,\qquad 
\frac{19}{15}\approx1.2667,\qquad 
\frac{17}{13}\approx1.3077
\]

From smallest to largest the order is  

\[
\frac{19}{15}<\frac{17}{13}<\frac{15}{11}.
\]

This matches choice **E**.

**ANSWER 1:** E  

---

**Problem 2:**  
Let \(B\) be the number of students who have both a dog and a cat.  
Using inclusion–exclusion for the 39 students:

\[
\text{total}= (\text{dog owners})+(\text{cat owners})-B
\]
\[
39 = 20 + 26 - B \;\Longrightarrow\; B = 20+26-39 = 7.
\]

So 7 students have both a dog and a cat.

**ANSWER 2:** A  

---

**Problem 3:**  
Start with \$100.  
First year loss of 15%: \(100\times0.85 = \$85\).  
Second year gain of 20%: \(85\times1.20 = \$102\).

The investment grew from \$100 to \$102, a net increase of \$2, i.e. a **2 % gain**.

**ANSWER 3:** D  

---

**Problem 4:**  
Wind‑chill formula:  

\[
\text{WC}= \text{air temp} - 0.7\times(\text{wind speed})
\]
\[
\text{WC}= 36 - 0.7(18)=36-12.6=23.4.
\]

The value closest to this is **23**.

**ANSWER 4:** B  

---

**Problem 5:**  
Let the total number of marbles be \(N\).

\[
\begin{aligned}
\text{red} &= N-6,\\
\text{green}&= N-8,\\
\text{blue} &= N-4.
\end{aligned}
\]

Since all marbles are either red, green, or blue:

\[
(N-6)+(N-8)+(N-4)=N \;\Longrightarrow\; 3N-18=N \;\Longrightarrow\; N=9.
\]

There are **9** marbles in the jar.

**ANSWER 5:** C  

---

**Problem 6:**  
Isabella needs a total of \(95\times4 = 380\) points.  
After three tests she has \(97+91+x = 188+x\) points, where \(x\) is her third‑test score.  
To still be able to reach 380, the required fourth‑test score is

\[
y = 380-(188+x)=192-x.
\]

She can achieve this only if \(0\le y\le100\).  
Thus \(192-x\le100 \;\Longrightarrow\; x\ge92\).  
The lowest possible integer score satisfying this is **92**.

**ANSWER 6:** B  

---

**Problem 7:**  
Let the tree heights be \(3k\) and \(4k\) (ratio \(3:4\)).  
The taller tree exceeds the shorter by \(4k-3k=k=16\) ft.  
Hence \(k=16\) and the taller tree’s height is \(4k=64\) ft.

**ANSWER 7:** B  

---

**Problem 8:**  
In a single‑elimination tournament each game eliminates exactly one team.  
With 16 teams, 15 teams must be eliminated to leave a champion, so **15 games** are played.

**ANSWER 8:** D  

---

**Problem 9:**  
We need triples of positive integers \((a,b,c)\) with \(a<b<c\) and \(abc=100\).

Factorizations of 100 (up to order) are:

\[
\begin{aligned}
&(1,1,100),\;(1,2,50),\;(1,4,25),\;(1,5,20),\;(1,10,10),\\
&(2,2,25),\;(2,5,10),\;(4,5,5).
\end{aligned}
\]

Only the triples with all three numbers distinct satisfy \(a<b<c\):

\[
(1,2,50),\;(1,4,25),\;(1,5,20),\;(2,5,10).
\]

Thus there are **4** possible ways.

**ANSWER 9:** E  

---

**Problem 10:**  
Three‑digit numbers run from 100 to 999.  
The smallest multiple of 13 is \(13\times8=104\).  
The largest multiple ≤ 999 is \(13\times76=988\).

Number of multiples:

\[
\frac{988-104}{13}+1 = \frac{884}{13}+1 = 68+1 = 69.
\]

So there are **69** three‑digit numbers divisible by 13.

**ANSWER 10:** C  

---

**Problem 11:**  
To maximize the difference \(a-b\) we choose the largest possible minuend and the smallest possible subtrahend from the set \(\{-16,-4,0,2,4,12\}\).

\[
\max(a-b)=12-(-16)=28.
\]

No other pair yields a larger value. Hence the largest difference is **28**.

**ANSWER 11:** D  

---

**Problem 12:**  
Boy: 15 miles in 3 h 30 min = 210 min → time per mile = \(210/15 = 14\) min.  
Old man: 10 miles in 4 h = 240 min → time per mile = \(240/10 = 24\) min.  

Difference = \(24-14 = 10\) minutes.

**ANSWER 12:** B

Problem 1:  
We need to order the fractions $ \frac{15}{11} $, $ \frac{19}{15} $, and $ \frac{17}{13} $ from least to greatest.  
Convert each fraction to a decimal:  
$ \frac{15}{11} \approx 1.3636 $  
$ \frac{19}{15} \approx 1.2667 $  
$ \frac{17}{13} \approx 1.3077 $  
Ordering: $ \frac{19}{15} < \frac{17}{13} < \frac{15}{11} $.  
The correct choice is E.  
ANSWER 1: E  

Problem 2:  
Total students = 39.  
Students with a dog = 20, with a cat = 26.  
Using the inclusion-exclusion principle:  
$ |D \cup C| = |D| + |C| - |D \cap C| $  
$ 39 = 20 + 26 - |D \cap C| $  
$ |D \cap C| = 7 $.  
ANSWER 2: A  

Problem 3:  
Initial investment = $100.  
After 15% loss: $ 100 \times 0.85 = 85 $.  
After 20% gain: $ 85 \times 1.20 = 102 $.  
Change: $ 102 - 100 = 2 $ (2% gain).  
ANSWER 3: D  

Problem 4:  
Wind chill formula: $ \text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed} $.  
Given: Air temperature = 36°F, Wind speed = 18 mph.  
Calculation: $ 36 - 0.7 \times 18 = 36 - 12.6 = 23.4 $.  
Closest value: 23.  
ANSWER 4: A (closest is 23, but 23.4 is closer to 23 than 28? Actually 23.4 - 23 = 0.4, 28 - 23.4 = 4.6, so 23 is closer. However, the choices are A.18, B.23, C.28, D.32, E.35. So B.23 is correct.)  
But in the calculation it's 23.4, which is closer to 23.  
ANSWER 4: B  

Problem 5:  
Let $ R $, $ G $, $ B $ be the number of red, green, and blue marbles.  
Given:  
$ G + B = 6 $ (all but red)  
$ R + B = 8 $ (all but green)  
$ R + G = 4 $ (all but blue)  
Adding all three: $ 2(R + G + B) = 18 $, so $ R + G + B = 9 $.  
ANSWER 5: C  

Problem 6:  
Goal average = 95 over 4 tests, so total points needed = $ 95 \times 4 = 380 $.  
First two tests: 97 + 91 = 188.  
Let third test score = $ x $, fourth test score = $ y $.  
$ 188 + x + y = 380 $ → $ x + y = 192 $.  
To minimize $ x $, maximize $ y $ (since $ y \leq 100 $).  
$ x = 192 - 100 = 92 $.  
ANSWER 6: B  

Problem 7:  
Let heights of trees be $ 3x $ and $ 4x $.  
Difference: $ 4x - 3x = 16 $ → $ x = 16 $.  
Taller tree: $ 4x = 64 $.  
ANSWER 7: B  

Problem 8:  
16 teams, single elimination.  
Number of games = number of teams - 1 = 15.  
ANSWER 8: D  

Problem 9:  
Find triples $ (a, b, c) $ with $ a < b < c $ and $ a \times b \times c = 100 $.  
Factorize 100: $ 100 = 2^2 \times 5^2 $.  
Possible triples:  
- $ (1, 4, 25) $: $ 1 \times 4 \times 25 = 100 $  
- $ (1, 5, 20) $: $ 1 \times 5 \times 20 = 100 $  
- $ (2, 5, 10) $: $ 2 \times 5 \times 10 = 100 $  
But $ a < b < c $, so only $ (1,4,25), (1,5,20), (2,5,10) $? Actually $ (2,5,10) $ is valid.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $, but $ 50 $ is larger.  
Actually, $ (1,4,25), (1,5,20), (2,5,10) $.  
But $ (1,1,100) $ invalid since $ a < b < c $.  
So 4? Wait, $ (1,4,25), (1,5,20), (2,5,10) $, and $ (1,2,50) $ but $ 1<2<50 $.  
But $ (1,2,50) $: $ 1 \times 2 \times 50 = 100 $.  
And $ (1,10,10) $ invalid.  
So 4 ways? But choices up to 4.  
List:  
- 1,4,25  
- 1,5,20  
- 2,5,10  
- 1,2,50  
But $ (1,1,100) $ invalid.  
So 4 ways.  
ANSWER 9: E  

Problem 10:  
First three-digit number divisible by 13: 104 (13×8).  
Last: 988 (13×76).  
Number of terms: $ 76 - 8 + 1 = 69 $.  
ANSWER 10: C  

Problem 11:  
Set: {−16, −4, 0, 2, 4, 12}.  
Largest difference: max - min = 12 - (-16) = 28.  
ANSWER 11: D  

Problem 12:  
As a boy: 15 miles in 3.5 hours → $ \frac{15}{3.5} \approx 4.2857 $ miles per hour → $ \frac{3.5}{15} \times 60 = 14 $ minutes per mile.  
As an old man: 10 miles in 4 hours → $ \frac{4}{10} \times 60 = 24 $ minutes per mile.  
Difference: $ 24 - 14 = 10 $ minutes.  
ANSWER 12: B

---
### Problem 1:
**What is asked**: Order the improper fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
**Setup**: All fractions can be rewritten as $1 + \text{(a remainder fraction)}$ to simplify comparison:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. So $\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$. Adding 1 to each preserves the order, so $\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$.
**Wrong choices ruled out**: A/B reverse the order, C incorrectly lists $\frac{17}{13}$ as the smallest, D mixes up the middle terms.
ANSWER 1: E

---
### Problem 2:
**What is asked**: Find how many of 39 students (all have at least one dog or cat) have both a dog and a cat, given 20 have dogs, 26 have cats.
**Setup**: Use the inclusion-exclusion principle for sets: $|D \cup C| = |D| + |C| - |D \cap C|$. All students have at least one pet, so $|D \cup C| = 39$. Let $x = |D \cap C|$ (students with both pets):
$39 = 20 + 26 - x \implies 39 = 46 - x \implies x =7$.
**Wrong choices ruled out**: B is the number of students with only a dog, E is the sum of dog and cat owners before subtracting overlaps, C/D are far too large.
ANSWER 2: A

---
### Problem 3:
**What is asked**: Calculate the overall percent change in a $100 investment that loses 15% in year 1, then gains 20% in year 2.
**Setup**: Calculate the final value of the investment:
- After year 1 (15% loss): $100 * (1 - 0.15) = \$85$
- After year 2 (20% gain): $85 * (1 + 0.20) = \$102$
The final value is $102, a 2% gain over the original $100.
**Wrong choices ruled out**: E is the tempting mistake of adding the percentages directly (-15% +20% =5% gain), B incorrectly claims a loss, A/C are incorrect magnitudes.
ANSWER 3: D

---
### Problem 4:
**What is asked**: Calculate wind chill for 36°F air temperature and 18mph wind, using the formula $\text{wind chill} = \text{air temp} - 0.7 * \text{wind speed}$.
**Setup**: Plug in values:
$0.7 * 18 = 12.6$, so wind chill = $36 -12.6 = 23.4$, which is closest to 23.
**Wrong choices ruled out**: A/C/D/E are all miscalculations from misapplying the formula.
ANSWER 4: B

---
### Problem 5:
**What is asked**: Find the total number of marbles in a jar of red, green, blue marbles, where "all but 6 are red, all but 8 are green, all but 4 are blue".
**Setup**: Let $R$=red, $G$=green, $B$=blue, $T=R+G+B$ (total marbles). The conditions translate to:
- $G+B =6$ (all but 6 are red)
- $R+B=8$ (all but 8 are green)
- $R+G=4$ (all but 4 are blue)
Add the three equations: $2(R+G+B) = 6+8+4=18 \implies T=9$. Verify: $R=3, G=1, B=5$, sum to 9 and satisfy all conditions.
**Wrong choices ruled out**: B/D/E are incorrect totals that fail the marble count conditions.
ANSWER 5: C

---
### Problem 6:
**What is asked**: Find the lowest possible third test score Isabella can have to still reach a 95 average on four 100-point tests, after scoring 97 and 91 on the first two.
**Setup**: Total points needed for a 95 average: $4*95=380$ total points. Sum of first two tests: $97+91=188$, so $380-188=192$ points needed from the last two tests. To minimize the third test, set the fourth test to its maximum possible value (100), so the third test is $192-100=92$.
**Wrong choices ruled out**: A (90) is too low to reach 192 total points from the last two tests, C/D/E are higher than the minimum possible score.
ANSWER 6: B

---
### Problem 7:
**What is asked**: Find the height of the taller tree, where two trees have heights in a 3:4 ratio, and the taller tree is 16ft higher than the shorter.
**Setup**: Let the heights be $3x$ (shorter) and $4x$ (taller). The height difference is $4x-3x=x=16$. The taller tree is $4x=4*16=64$ft.
**Wrong choices ruled out**: A is the height of the shorter tree, C/D/E are far too large.
ANSWER 7: B

---
### Problem 8:
**What is asked**: Find the number of games needed to determine a winner in a 16-team elimination tournament (each game eliminates 1 losing team).
**Setup**: To get 1 winner, 15 of the 16 teams must be eliminated. Since each game eliminates exactly 1 team, 15 games are required. Adding round-by-round (8 first-round games +4 quarterfinals +2 semifinals +1 final =15) confirms this.
**Wrong choices ruled out**: A is the number of tournament rounds, not games, E is the number of teams, B/C are too low.
ANSWER 8: D

---
### Problem 9:
**What is asked**: Count the number of sets of three positive integers $a<b<c$ whose product is 100.
**Setup**: First factorize $100=2^25^2$, list all strictly increasing triples that multiply to 100:
1. $(1,2,50)$: $1*2*50=100$
2. $(1,4,25)$: $1*4*25=100$
3. $(1,5,20)$: $1*5*20=100$
4. $(2,5,10)$: $2*5*10=100$
These are all the strictly increasing valid triples, so there are 4 ways.
**Wrong choices ruled out**: A/B/C/D underestimate the number of valid triples.
ANSWER 9: E

---
### Problem 10:
**What is asked**: Find how many three-digit numbers are divisible by 13.
**Setup**: Find the first and last three-digit multiples of 13:
- Smallest three-digit multiple: $13*8=104$
- Largest three-digit multiple: $13*76=988$ ($13*77=1001$ is four-digit)
Count the number of multiples from 8 to 76 inclusive: $76-8+1=69$.
**Wrong choices ruled out**: B/D/E are miscalculations of the count of multiples.
ANSWER 10: C

---
### Problem 11:
**What is asked**: Find the largest difference possible by subtracting two numbers from the set $\{-16,-4,0,2,4,12\}$.
**Setup**: To maximize the value of $x-y$ (the difference from subtracting $y$ from $x$), use the largest $x$ and smallest $y$ in the set: $12 - (-16)=28$.
**Wrong choices ruled out**: E is the product of the two extreme values, C is far too small, A/B are not maximum.
ANSWER 11: D

---
### Problem 12:
**What is asked**: Find how many minutes longer it takes Cheenu to travel 1 mile now (as an old man) than when he was a boy, given he ran 15 miles in 3h30m 

### Problem 1:
We need to order the fractions $\frac{15}{11}$, $\frac{19}{15}$, $\frac{17}{13}$ from least to greatest.
First, rewrite each fraction as $1 + \frac{4}{\text{denominator}}$:
- $\frac{15}{11} = 1 + \frac{4}{11}$
- $\frac{19}{15} = 1 + \frac{4}{15}$
- $\frac{17}{13} = 1 + \frac{4}{13}$
For fractions with the same numerator, a larger denominator means a smaller fraction. Thus:
$\frac{4}{15} < \frac{4}{13} < \frac{4}{11}$
Adding 1 to each term preserves the order:
$\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$
ANSWER 1: E
---
### Problem 2:
We use the inclusion-exclusion principle for sets. Let $D$ = students with a dog, $C$ = students with a cat. We know:
- $|D \cup C| = 39$ (all students have at least one pet)
- $|D| = 20$, $|C| = 26$
The formula is $|D \cup C| = |D| + |C| - |D \cap C|$, where $|D \cap C|$ is the number of students with both pets. Substitute values:
$39 = 20 + 26 - |D \cap C|$
$39 = 46 - |D \cap C|$
$|D \cap C| = 46 - 39 = 7$
ANSWER 2: A
---
### Problem 3:
Calculate the final value of the $100 investment:
1. First year: 15% loss, so remaining amount = $100 \times (1 - 0.15) = 100 \times 0.85 = \$85$
2. Second year: 20% gain on the remaining $85, so final amount = $85 \times (1 + 0.20) = 85 \times 1.2 = \$102$
The net change is $102 - 100 = \$2$, which is a 2% gain on the original $100.
ANSWER 3: D
---
### Problem 4:
Use the wind chill formula: $\text{Wind chill} = \text{Air temperature} - 0.7 \times \text{Wind speed}$. Substitute air temperature = 36°F, wind speed = 18 mph:
$\text{Wind chill} = 36 - 0.7 \times 18 = 36 - 12.6 = 23.4$
The closest value to 23.4 is 23.
ANSWER 4: B
---
### Problem 5:
Let $R$ = red marbles, $G$ = green marbles, $B$ = blue marbles. Translate the problem statements into equations:
1. "All but 6 are red": Non-red marbles = 6, so $G + B = 6$
2. "All but 8 are green": Non-green marbles = 8, so $R + B = 8$
3. "All but 4 are blue": Non-blue marbles = 4, so $R + G = 4$
Add all three equations:
$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$
$2(R + G + B) = 18$
$R + G + B = 9$
Total marbles = 9.
ANSWER 5: C
---
### Problem 6:
Isabella needs an average of 95 over 4 tests, so total required points = $4 \times 95 = 380$. Her first two test scores sum to $97 + 91 = 188$. Let her third test score be $x$, fourth test score be $y$ (maximum 100, since tests are 100-point). We need:
$188 + x + y = 380 \implies x + y = 192$
To minimize $x$, maximize $y = 100$. Thus:
$x = 192 - 100 = 92$
ANSWER 6: B
---
### Problem 7:
Let the heights of the two trees be $3k$ and $4k$ (ratio 3:4). The taller tree is $4k$, the shorter is $3k$. The difference in height is $4k - 3k = k = 16$ feet. Thus, the taller tree is $4k = 4 \times 16 = 64$ feet.
ANSWER 7: B
---
### Problem 8:
In a single-elimination tournament, every game eliminates exactly 1 team. To determine a winner from 16 teams, we need to eliminate 15 teams, so 15 games are played. Verify by round:
- Round 1: 16 teams → 8 games, 8 winners
- Round 2: 8 teams → 4 games, 4 winners
- Round 3: 4 teams → 2 games, 2 winners
- Final: 2 teams → 1 game
Total: $8 + 4 + 2 + 1 = 15$ games.
ANSWER 8: D
---
### Problem 9:
We need distinct positive integers $a < b < c$ with $a \times b \times c = 100$. Factorize 100: $100 = 2^2 \times 5^2$. List all valid triplets:
1. $1 \times 2 \times 50 = 100$
2. $1 \times 4 \times 25 = 100$
3. $1 \times 5 \times 20 = 100$
4. $2 \times 5 \times 10 = 100$
Triplets with duplicates (e.g., $1 \times 10 \times 10$, $2 \times 2 \times 25$) are invalid because $a < b < c$ requires distinct values. There are 4 valid triplets.
ANSWER 9: E
---
### Problem 10:
Three-digit numbers range from 100 to 999. Find the smallest and largest three-digit multiples of 13:
- Smallest: $13 \times 8 = 104$ (since $13 \times7=91$ is two-digit)
- Largest: $13 \times76 = 988$ (since $13 \times77=1001$ is four-digit)
The number of multiples is $\frac{\text{Largest} - \text{Smallest}}{13} + 1 = \frac{988 - 104}{13} + 1 = \frac{884}{13} + 1 = 68 + 1 = 69$.
ANSWER 10: C
---
### Problem 11:
The largest difference between two numbers in a set is (maximum value) - (minimum value). For the set $\{-16, -4, 0, 2, 4, 12\}$:
- Maximum = 12, Minimum = -16
- Largest difference = $12 - (-16) = 12 + 16 = 28$
ANSWER 11: D
---
### Problem 12:
Calculate time per mile for boy and old man:
- Boy: 3 hours 30 minutes = $3 \times 60 + 30 = 210$ minutes for 15 miles. Time per mile = $\frac{210}{15} = 14$ minutes.
- Old man: 4 hours = $4 \times 60 = 240$ minutes for 10 miles. Time per mile = $\frac{240}{10} = 24$ minutes.
Difference: $24 - 14 = 10$ minutes longer per mile.
ANSWER 12: B