🔬 Model Lab

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]

Problem 1:
We are asked for the average weight of all 10 children together.

The average of a group is (total weight) ÷ (number of people). So find totals.

For the 6 boys: average 150 pounds.
Total boys' weight = 6 × 150 = 900 pounds.

For the 4 girls: average 120 pounds.
Total girls' weight = 4 × 120 = 480 pounds.

Combined total = 900 + 480 = 1380 pounds.
Number of children = 6 + 4 = 10.

Average = 1380 ÷ 10 = 138 pounds.

Note the tempting wrong choice A 135 is just (150+120)/2, the simple average of the two averages. That is wrong because there are not equal numbers of boys and girls. We must weight by 6 and 4.
ANSWER 1: C

Problem 2:
We start with a 4 inch by 4 inch square. Fold it in half vertically.

After folding we have a double-layer rectangle 2 inches wide by 4 inches tall. The fold is on one long side.

Now cut both layers with a cut parallel to the fold, i.e. a vertical cut, so as to halve it. Cut in the middle, 1 inch from the edge.

Call the fold side x=2, open side x=4. Cut at x=3.

- The piece containing the fold stays joined at the fold. Each layer is 1 inch wide by 4 tall. Joined together it unfolds to 2 inches wide by 4 tall. This is the large rectangle: 2 by 4.
- The other piece has no fold, so it falls into two separate pieces, each 1 inch wide by 4 tall. These are the two small rectangles: 1 by 4.

Perimeter of large: 2 × (2+4) = 12 inches.
Perimeter of one small: 2 × (1+4) = 10 inches.

Ratio small : large = 10 : 12 = 10/12 = 5/6.

Choice B 1/2 etc. would come from confusing area or side lengths.
ANSWER 2: E

Problem 3:
We need 4-digit numbers ABCD with:
A odd, B even, all four digits different.

Count choices for each position.

A is the leftmost digit of a 4-digit number, so A ≠ 0. Odd digits: 1,3,5,7,9. That is 5 choices.

B is even: 0,2,4,6,8. That is 5 choices. Since A is odd and B is even, B is automatically different from A, so no reduction. 5 choices remain.

C is the third digit: no odd/even restriction, only must be different from A and B. Two digits used, so 10-2 = 8 choices remain.

D is the fourth digit: must be different from A,B,C. Three digits used, so 10-3 = 7 choices remain.

Total numbers = 5 × 5 × 8 × 7 = 25 × 56 = 1400.

Tempting errors are forgetting A cannot be zero or miscounting B.
ANSWER 3: B

Problem 4:
Each starts at $100. Let A,B,C be finals.

AA: up 20% then down 20%.
After year 1: 100 × 1.20 = 120.
After year 2: 120 × 0.80 = 96. So A = 96.

BB: down 25% then up 25%.
After year 1: 100 × 0.75 = 75.
After year 2: 75 × 1.25 = 93.75. So B = 93.75.

CC: unchanged: C = 100.

Hence 93.75 < 96 < 100, i.e. B < A < C.

Note A = B = C is wrong: a % up then same % down does not return to start, it loses. And B loses even more: 0.75×1.25 = 0.9375 < 0.96 = 1.2×0.8.
ANSWER 4: E

Problem 5:
Original $180. Sale is 50% below original, then Saturday is 20% off the sale price.

Sale price = 50% off means pay 50%:
180 × 0.50 = $90.

Saturday: 20% off sale means pay 80% of sale:
90 × 0.80 = $72.

The tempting choice A $54 adds the discounts: 50%+20%=70% off, 30% of 180 = 54. That is wrong because the second discount is on the already reduced price, not the original.
ANSWER 5: B

Problem 6:
We need (total miles driven) ÷ (total gallons burned).

Miles: end odometer minus start:
57,060 − 56,200 = 860 miles.

Gallons: The driver fills to full each time.
- 6 gal at start just tops the tank before the trip; it is not burned on the trip.
- Drive 56,200 to 56,560, then add 12 gal to fill again. So 12 gal were burned on the first leg.
- Drive 56,560 to 57,060, then add 20 gal to fill again. So 20 gal were burned on the second leg.

Total burned = 12 + 20 = 32 gallons.

Average mpg = 860 ÷ 32 = 26.875 ≈ 26.9 to the nearest tenth.

Tempting choice B 22.6 comes from using 6+12+20=38 gallons: 860/38≈22.6. That wrongly includes the initial 6 gallons which were not used on the trip, while correctly including the final fill which replaces gas used.
ANSWER 6: D

Problem 7:
Need 5x + 10y + 25z = 710 cents with 0 ≤ x,y,z ≤ 20, maximize N = x+y+z, where x,y,z are numbers of 5c,10c,25c stamps.

Divide by 5: x + 2y + 5z = 142.

Then N = x+y+z = 142 − y − 4z. So to maximize N, minimize S = y + 4z.

Also x = 142 − 2y − 5z must satisfy 0 ≤ x ≤ 20, i.e.
122 ≤ 2y + 5z ≤ 142.

To make S small, want z small. But y ≤ 20 limits how small z can be.

Try z=16: need 2y+80 ≥122 → y≥21, impossible since y≤20.
So z≥17.

For z=17: x = 142−2y−85 = 57−2y. Need 0≤x≤20 → 37≤2y≤57 → y=19 or 20 (with y≤20). To minimize S=y+68, take smallest y=19. Then x=57−38=19. This is allowed: ≤20.

This gives: x=19, y=19, z=17. Check value: 19×5+19×10+17×25 =95+190+425=710. Number of stamps =19+19+17=55.

Any z≥18 gives S = y+4z ≥ ... For z=18, need y≥16, so S≥88; our S=19+68=87 is smaller. Larger z gives even larger S. Hence 55 is maximal.

Simply using all twenty 5c stamps forces z=18, y=16, N=54, so D 54 is tempting but not maximal.
ANSWER 7: E

Problem 8:
Use digits 2,4,5,7 once each. 4! = 24 numbers. Sort smallest to largest.

First digit determines blocks of 3! = 6:
1-6 start with 2,
7-12 start with 4,
13-18 start with 5,
19-24 start with 7.

17th is in the 13-18 block, so starts with 5.

Fix first digit 5. Remaining {2,4,7}. Second digit determines blocks of 2! = 2:
13-14: 52--,
15-16: 54--,
17-18: 57--.

So 17th starts 57. Remaining digits {2,4}:
smallest is 5724, then 5742.

Thus 17th = 5724, 18th = 5742. Choice C is the 18th.
ANSWER 8: B

Problem 9:
The product is
(3/2)·(4/3)·(5/4)·(6/5)· … ·(a/b) = 9.

This telescopes: numerators cancel denominators of the next fraction.

Indeed the pattern is k/(k-1). So the last term must be a/(a-1), i.e. b = a−1.

Product = a/2, because all intermediate numbers cancel, leaving last numerator a over first denominator 2.

Set a/2 = 9 → a = 18. Then b = 17.

Sum a+b = 18+17 = 35.
ANSWER 9: D

Problem 10:
In triangle ABC, angle B = 50°. AD bisects angle BAC, DC bisects angle BCA. Find angle ADC.

Let angle A = BAC, angle C = BCA.
A + B + C = 180°, so A + C = 180° − 5

Here are the step-by-step solutions for each problem:

---

### Problem 1:
**Restatement:** We are given that 6 boys have an average weight of 150 pounds and 4 girls have an average weight of 120 pounds. We need to find the average weight of all 10 children.

**Step-by-step solution:**
1. Find the total weight of the 6 boys:
   $$\text{Total weight of boys} = 6 \times 150 = 900\text{ pounds}$$
2. Find the total weight of the 4 girls:
   $$\text{Total weight of girls} = 4 \times 120 = 480\text{ pounds}$$
3. Add these to get the total weight of all 10 children:
   $$\text{Total weight} = 900 + 480 = 1380\text{ pounds}$$
4. Calculate the average weight of the 10 children:
   $$\text{Average weight} = \frac{1380}{10} = 138\text{ pounds}$$

A common mistake is simply averaging the two given averages: $\frac{150 + 120}{2} = 135$ (Choice A), which is incorrect because there are more boys than girls.

ANSWER 1: C

---

### Problem 2:
**Restatement:** A $4 \times 4$-inch square sheet of paper is folded in half vertically, creating a folded piece measuring 2 inches by 4 inches. Both layers are then cut in half vertically (parallel to the fold). We need to find the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle.

**Step-by-step solution:**
1. The fold is along the vertical centerline. The folded paper has a width of $2$ inches and a height of $4$ inches.
2. Cutting in half parallel to the fold cuts along the vertical line $1$ inch from the crease:
   - The piece containing the crease, when unfolded, forms a single large rectangle with dimensions $2\text{ inches} \times 4\text{ inches}$.
   - The other cut piece consists of two separate, identical small strips, each with dimensions $1\text{ inch} \times 4\text{ inches}$.
3. Calculate the perimeters:
   - Perimeter of one small rectangle $= 2 \times (1 + 4) = 10\text{ inches}$.
   - Perimeter of the large rectangle $= 2 \times (2 + 4) = 12\text{ inches}$.
4. Find the ratio of the perimeter of a small rectangle to that of the large rectangle:
   $$\text{Ratio} = \frac{10}{12} = \frac{5}{6}$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** We want to find the number of four-digit whole numbers where the first (leftmost) digit is odd, the second digit is even, and all four digits are distinct.

**Step-by-step solution:**
1. **Leftmost (thousands) digit:** Must be odd, so it can be any of $\{1, 3, 5, 7, 9\}$. There are $5$ possibilities.
2. **Second (hundreds) digit:** Must be even, so it can be any of $\{0, 2, 4, 6, 8\}$. Since all odd digits and even digits are disjoint, none of these overlap with the first digit. Thus, there are $5$ possibilities.
3. **Third (tens) digit:** Can be any digit from $0$ to $9$, except the two digits already selected. There are $10 - 2 = 8$ possibilities.
4. **Fourth (units) digit:** Can be any digit from $0$ to $9$, except the three digits already chosen. There are $10 - 3 = 7$ possibilities.
5. Multiply the number of choices for each position:
   $$\text{Total} = 5 \times 5 \times 8 \times 7 = 25 \times 56 = 1400$$

ANSWER 3: B

---

### Problem 4:
**Restatement:** Penni invests $\$100$ into three stocks: $AA$, $BB$, and $CC$. We need to track their values over two years and order their final values $A$, $B$, and $C$.

**Step-by-step solution:**
1. **Company $AA$:**
   - After Year 1 (+20%): $100 \times 1.20 = \$120$.
   - After Year 2 (-20% from new value): $120 \times 0.80 = \$96$.
   - So, $A = 96$.
2. **Company $BB$:**
   - After Year 1 (-25%): $100 \times 0.75 = \$75$.
   - After Year 2 (+25% from new value): $75 \times 1.25 = \$93.75$.
   - So, $B = 93.75$.
3. **Company $CC$:**
   - Unchanged in both years, so $C = 100$.
4. Comparing the values:
   $$93.75 < 96 < 100 \implies B < A < C$$

Note: Many people mistakenly think a $+20\%$ followed by a $-20\%$ brings the stock back to the original $\$100$, which would give $A = B = C$. However, the percentage changes are applied to different base amounts.

ANSWER 4: E

---

### Problem 5:
**Restatement:** A coat originally costs $\$180$. The outlet has a base discount of 50%, and an extra 20% discount off the sale price on Saturdays. We want to find the Saturday price.

**Step-by-step solution:**
1. Calculate the regular sale price (50% off the original):
   $$\text{Sale Price} = 180 \times (1 - 0.50) = 180 \times 0.50 = \$90$$
2. Apply the Saturday discount of 20% to the sale price:
   $$\text{Saturday Price} = 90 \times (1 - 0.20) = 90 \times 0.80 = \$72$$

A common error is adding the percentages together ($50\% + 20\% = 70\%$ off), which would give $180 \times 0.30 = \$54$ (Choice A). The problem explicitly states the additional discount is taken *off the sale price*.

ANSWER 5: B

---

### Problem 6:
**Restatement:** The odometer reads 56,200 at the start when filled with 6 gallons. It is filled with 12 gallons at 56,560 miles, and with 20 gallons at the end of the trip at 57,060 miles. We need the average miles-per-gallon (mpg) for the entire trip.

**Step-by-step solution:**
1. **Total distance traveled:**
   $$\text{Distance} = 57{,}060 - 56{,}200 = 860\text{ miles}$$
2. **Total gasoline consumed:**
   The 6 gallons added at the start merely topped off the tank to make it full before the trip began. The gasoline consumed during the trip is replaced by the subsequent fill-ups:
   - 12 gallons to fill back up during the trip.
   - 20 gallons to fill back up at the end of the trip.
   $$\text{Total gas used} = 12 + 20 = 32\text{ gallons}$$
3. **Average miles-per-gallon:**
   $$\text{MPG} = \frac{860}{32} = \frac{215}{8} = 26.875\text{ mpg}$$
4. Rounding to the nearest tenth gives $26.9$ mpg.

ANSWER 6: D

---

### Problem 7:
**Restatement:** Nicolas has twenty 5¢ stamps, twenty 10¢ stamps, and twenty 25¢ stamps. We want to find the maximum number of stamps he can use to make exactly $\$7.10$ (710¢).

**Step-by-step solution:**
1. The total value of all 60 stamps is:
   $$20 \times (5 + 10 + 25) = 20 \times 40 = 800\t

# Problem 1

**Restate:** 6 boys average 150 lb, 4 girls average 120 lb; find the average of all 10.

**Total weights:**
- Boys: 6 × 150 = 900 lb
- Girls: 4 × 120 = 480 lb
- Combined: 900 + 480 = 1380 lb

**Average:** 1380 ÷ 10 = 138 lb.

(The tempting wrong answer is 135, the plain average of 150 and 120 — but there are more boys than girls, so the mean must be pulled *above* 135.)

ANSWER 1: C

# Problem 2

**Restate:** A 4×4 square is folded in half vertically (making a 2×4 double layer). Then both layers are cut in half *parallel to the fold*, i.e., along a line 1 inch from the fold.

The cut produces two stacks:
- The stack containing the fold: two layers of 1×4 joined at the fold, which unfolds into one **2×4 large rectangle**.
- The other stack: two separate **1×4 small rectangles**.

**Perimeters:**
- Small: 2(1 + 4) = 10 inches
- Large: 2(2 + 4) = 12 inches

**Ratio:** 10/12 = 5/6.

ANSWER 2: E

# Problem 3

**Restate:** Count 4-digit numbers with first digit odd, second digit even, all four digits distinct.

- First digit odd: 1, 3, 5, 7, 9 → 5 choices (none is 0, so it's a valid leading digit).
- Second digit even: 0, 2, 4, 6, 8 → 5 choices (can't duplicate the first digit, but odd ≠ even automatically, so all 5 work).
- Third digit: any of the remaining 10 − 2 = 8 digits.
- Fourth digit: any of the remaining 7 digits.

**Total:** 5 × 5 × 8 × 7 = 1400.

(2500 = 5·5·10·10 forgets "all different"; 1120 = 5·4·8·7 wrongly removes a choice from the even digit.)

ANSWER 3: B

# Problem 4

**Restate:** Each stock starts at \$100; track two years of changes.

- **AA:** 100 × 1.20 = 120, then 120 × 0.80 = **96**
- **BB:** 100 × 0.75 = 75, then 75 × 1.25 = **93.75**
- **CC:** unchanged = **100**

So 93.75 < 96 < 100, i.e. **B < A < C**.

(The trap is thinking a 20% rise then 20% fall returns to \$100 — it doesn't; both A and B end below C, and B ends lowest.)

ANSWER 4: E

# Problem 5

**Restate:** 50% off, then an extra 20% off the sale price, on \$180.

- Sale price: 180 × 0.50 = \$90
- Saturday price: 90 × 0.80 = 90 − 18 = **\$72**

(\$90 forgets the second discount; \$54 wrongly takes 50% + 20% = 70% off the original.)

ANSWER 5: B

# Problem 6

**Restate:** Find miles per gallon over the whole trip.

**Miles driven:** 57,060 − 56,200 = 860 miles.

**Gas used during the trip:** The 6 gallons at the start just fills the tank before driving (it isn't consumed yet as a measured amount). Each later fill-up replaces exactly the gas burned since the previous fill:
- 12 gallons (burned over the first 360 miles)
- 20 gallons (burned over the last 500 miles)

Total gas used = 12 + 20 = 32 gallons.

**MPG:** 860 ÷ 32 = 26.875 ≈ **26.9**.

(Including the initial 6 gallons gives 860/38 ≈ 22.6 — the classic trap.)

ANSWER 6: D

# Problem 7

**Restate:** With at most 20 each of 5¢, 10¢, 25¢ stamps, total exactly 710¢, maximize the number of stamps.

Let $a,b,c$ = numbers of 5¢, 10¢, 25¢ stamps ($0\le a,b,c\le 20$):
$$5a+10b+25c=710 \Rightarrow a+2b+5c=142$$

Solve for $c=\dfrac{142-a-2b}{5}$, so the stamp count is
$$a+b+c = a+b+\frac{142-a-2b}{5}=\frac{4a+3b+142}{5}.$$

So we want $4a+3b$ as large as possible, with $142-a-2b$ divisible by 5 (i.e. $a+2b\equiv 2 \pmod 5$) and $c\le 20$.

Trying the largest values of $4a+3b$ in order: (20,20)→140, (20,19)→137, (19,20)→136, (20,18)→134, (19,19)→133, (18,20)→132, (20,17)→131, (19,18)→130, (18,19)→129 — none satisfy $a+2b\equiv2\pmod5$. Next: $(a,b)=(20,16)$ gives $a+2b=52\equiv 2$ ✓, with $4a+3b=128$.

Then $c=\frac{142-52}{5}=18\le 20$, and the count is $\frac{128+142}{5}=54$.

Check: 20(5) + 16(10) + 18(25) = 100 + 160 + 450 = 710¢ ✓, using 20+16+18 = **54** stamps. (55 is impossible, as the count $\frac{4a+3b+142}{5}$ can't reach 55 under the constraints.)

ANSWER 7: D

# Problem 8

**Restate:** All 24 permutations of digits 2, 4, 5, 7 listed increasingly; find the 17th.

Each leading digit accounts for 3! = 6 numbers:
- Start with 2: positions 1–6
- Start with 4: positions 7–12
- Start with 5: positions 13–18
- Start with 7: positions 19–24

So the 17th is the 5th number beginning with 5. The remaining digits 2, 4, 7 in increasing order: 247, 274, 427, 472, **724**, 742. The 5th is 724.

Thus the number is **5724**.

ANSWER 8: B

# Problem 9

**Restate:** $\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac{a}{b}=9$; find $a+b$.

Each fraction is $\frac{n+1}{n}$, so consecutive factors telescope:
$$\frac32\cdot\frac43\cdot\frac54\cdots\frac{a}{a-1}=\frac{a}{2}$$

Set $\frac{a}{2}=9 \Rightarrow a=18$, and $b=a-1=17$.

**Sum:** 18 + 17 = 35.

ANSWER 9: D

# Problem 10

**Restate:** In triangle ABC, ∠ABC = 50°, and AD, CD bisect angles A and C (so D is where the two bisectors meet). Find ∠ADC.

Angles of the triangle: ∠A + ∠C = 180° − 50° = 130°.

The bisectors give ∠DAC = ½∠A and ∠DCA = ½∠C, so
$$\angle DAC+\angle DCA=\tfrac12(\angle A+\angle C)=\tfrac12(130°)=65°.$$

In triangle ADC:
$$\angle ADC = 180° - 65° = 115°.$$

(Equivalently, $\angle ADC = 90° + \tfrac12\angle B = 90° + 25° = 115°$. Note 90° would require the bisected angles to sum to 90°, and 125° comes from mistakenly using 180° − 50° − ... incorrectly.)

ANSWER 10: C

Problem 1:

We are asked for the average weight of all 10 children.

The 6 boys have average weight \(150\) pounds, so their total weight is

\[
6\cdot 150=900
\]

The 4 girls have average weight \(120\) pounds, so their total weight is

\[
4\cdot 120=480
\]

Thus, the total weight of all 10 children is

\[
900+480=1380
\]

The average is

\[
\frac{1380}{10}=138
\]

It is not correct simply to average \(150\) and \(120\), since there are more boys than girls.

\[
\boxed{138\text{ pounds}}
\]

ANSWER 1: C

Problem 2:

The original square is \(4\) inches by \(4\) inches.

After folding vertically, the paper has dimensions \(2\) inches by \(4\) inches. Cutting parallel to the fold divides the folded paper into two pieces that are each \(1\) inch by \(4\) inches.

When unfolded, this produces:

- one large middle rectangle measuring \(2\) by \(4\),
- two small rectangles measuring \(1\) by \(4\).

The perimeter of a small rectangle is

\[
2(1+4)=10
\]

The perimeter of the large rectangle is

\[
2(2+4)=12
\]

Therefore the desired ratio is

\[
\frac{10}{12}=\frac56
\]

The ratio is not \(\frac12\), because the small rectangle has the same height as the large rectangle; only its width is half as large.

\[
\boxed{\frac56}
\]

ANSWER 2: E

Problem 3:

We need four-digit numbers such that:

- the first digit is odd,
- the second digit is even,
- all four digits are different.

There are \(5\) choices for the first digit:

\[
1,3,5,7,9
\]

There are \(5\) choices for the second digit:

\[
0,2,4,6,8
\]

Since the first digit is odd and the second is even, they cannot be equal.

For the third digit, two digits have already been used, so there are

\[
10-2=8
\]

choices.

For the fourth digit, three digits have been used, so there are

\[
10-3=7
\]

choices.

Thus the total number is

\[
5\cdot 5\cdot 8\cdot 7=1400
\]

We do not subtract a choice for leading zero, since the first digit must be odd and therefore cannot be zero.

\[
\boxed{1400}
\]

ANSWER 3: B

Problem 4:

Each stock begins at \(\$100\).

For AA:

After rising \(20\%\),

\[
100(1.20)=120
\]

Then it drops \(20\%\):

\[
120(0.80)=96
\]

So,

\[
A=96
\]

For BB:

After dropping \(25\%\),

\[
100(0.75)=75
\]

Then it rises \(25\%\):

\[
75(1.25)=93.75
\]

So,

\[
B=93.75
\]

For CC, the value never changes:

\[
C=100
\]

Therefore,

\[
B<A<C
\]

A \(20\%\) increase followed by a \(20\%\) decrease does not return a stock to its original value, because the decrease is taken from the larger amount.

\[
\boxed{B<A<C}
\]

ANSWER 4: E

Problem 5:

The original price is \(\$180\).

First, the store takes \(50\%\) off:

\[
180(0.50)=90
\]

So the sale price is \(\$90\).

On Saturday, there is an additional \(20\%\) discount off the sale price:

\[
90(0.80)=72
\]

Therefore the Saturday price is

\[
\boxed{\$72}
\]

It is not \(\$54\), since that would result from treating the discounts as though they were both taken directly from the original price.

ANSWER 5: B

Problem 6:

The trip began at mileage

\[
56,\!200
\]

and ended at mileage

\[
57,\!060
\]

So the total distance traveled was

\[
57,\!060-56,\!200=860\text{ miles}
\]

The first 6 gallons were put in at the beginning, so they establish a full tank. The gasoline used during the trip is measured by the later fill-ups:

\[
12+20=32\text{ gallons}
\]

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875
\]

Rounded to the nearest tenth,

\[
26.9
\]

We do not include the initial 6 gallons, because those gallons were already in the tank at the start of the measured trip.

\[
\boxed{26.9}
\]

ANSWER 6: D

Problem 7:

Let \(x,y,z\) be the numbers of 5-cent, 10-cent, and 25-cent stamps used.

We want

\[
5x+10y+25z=710
\]

Dividing by \(5\),

\[
x+2y+5z=142
\]

We want as many stamps as possible.

Try using \(55\) stamps. The following choice works:

\[
x=19,\qquad y=19,\qquad z=17
\]

The number of stamps is

\[
19+19+17=55
\]

Their value is

\[
19(5)+19(10)+17(25)
\]

\[
95+190+425=710
\]

So 55 stamps can be used.

Could 56 stamps be used? With 56 stamps, the equation would be

\[
x+y+z=56
\]

and

\[
x+2y+5z=142
\]

Subtracting the first equation from the second gives

\[
y+4z=86
\]

From this,

\[
y=86-4z
\]

Also,

\[
x=56-y-z=56-(86-4z)-z=3z-30
\]

Since \(y\le20\),

\[
86-4z\le20
\]

\[
z\ge17
\]

But if \(z\ge17\), then

\[
x=3z-30\ge21
\]

which exceeds the allowed 20 five-cent stamps. So 56 stamps are impossible.

Therefore the greatest possible number is

\[
\boxed{55}
\]

ANSWER 7: E

Problem 8:

There are \(4!=24\) arrangements of the digits \(2,4,5,7\).

For each possible first digit, there are \(3!=6\) arrangements of the remaining digits.

Thus:

- Positions \(1\) through \(6\) begin with \(2\).
- Positions \(7\) through \(12\) begin with \(4\).
- Positions \(13\) through \(18\) begin with \(5\).
- Positions \(19\) through \(24\) begin with \(7\).

The 17th number begins with \(5\). It is the 5th number among those beginning with \(5\).

Arrange \(2,4,7\) in order:

\[
247,\ 274,\ 427,\ 472,\ 724,\ 742
\]

The 5th arrangement is \(724\), so the number is

\[
5724
\]

Thus the number in the 17th position is

\[
\boxed{5724}
\]

ANSWER 8: B

Problem 9:

The product is intended as

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots\frac ab=9
\]

This is a telescoping product:

\[
\frac32\cdot\frac43\cdot\frac54\cdot\frac65\cdots
\]

The \(3,4,5,\ldots\) terms cancel, leaving

\[
\frac a2=9
\]

Therefore,

\[
a=18
\]

The final fraction is \(\frac{18}{17}\), so

\[
b=17
\]

Hence,

\[
a+b=18+17=35
\]

\[
\boxed{35}
\]

ANSWER 9: D

Problem 10:

In triangle \(ABC\),

\[
\angle ABC=50^\circ
\]

Therefore the other two angles have total measure

\[
\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ
\]

Since \(AD\) bisects \(\angle BAC\),

\[
\angle CAD=\frac12\angle BAC
\]

Since \(DC\) bisects \(\angle BCA\),

\[
\angle ACD=\frac12\angle BCA
\]

In triangle \(ADC\),

\[
\angle ADC
=180^\circ-\angle CAD-\angle ACD
\]

\[
=180^\circ-\frac12(\angle BAC+\angle BCA)
\]