🔬 Model Lab

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D

Problem 1:
We are asked for the final price after a discount and then tax.

Original price is 80 dollars.
25% off means pay 100% - 25% = 75% of original.
Sale price = 0.75 × 80 = 60 dollars.

Tempting mistake: subtract 25 dollars to get 55. No, 25% means percent, not dollars.

Then 10% tax is added to the sale price, not the original price.
Tax = 10% of 60 = 6 dollars.
Total = 60 + 6 = 66 dollars. Equivalently 1.10 × 60 = 66.

Check: A 45 is far too low. B 52 would be 35% off total. C 54 would be adding tax incorrectly. E 68 would be adding 10% of 80.

ANSWER 1: D

Problem 2:
We need total time = time on coastal road + time on highway.

Given time on coastal road = 30 minutes for 10 miles.
Let coastal speed = v. Then v = 10 miles / 30 min.

Highway speed is three times as fast: 3v.
Highway distance is 50 miles.

Time = distance / speed, so:
Highway time = 50 / (3v).

From coastal: 10 / v = 30, so 1/v = 3 min per mile.
Thus highway time = 50 / 3 × (1/v) = 50/3 × 3 = 50 minutes.

Total = 30 + 50 = 80 minutes.

Alternative check: Highway is 5 times distance but 3 times speed, so takes 5/3 as long as coastal: 30 × 5/3 = 50.

ANSWER 2: C

Problem 3:
We are told w,x,y,z are whole numbers, so 0,1,2,...
And 2^w · 3^x · 5^y · 7^z = 588.
Find 2w + 3x + 5y + 7z.

Factor 588 into primes.
588 = 58.8 × 10, but do integer: 588 = 2 × 294 = 2 × 2 × 147 = 4 × 147.
147 = 3 × 49 = 3 × 7 × 7.
So 588 = 2^2 × 3^1 × 5^0 × 7^2.

By uniqueness of prime factorization, exponents must match:
w = 2, x = 1, y = 0, z = 2. Note y=0 is allowed because whole numbers include 0, and 5^0 = 1.

Now compute linear expression:
2w + 3x + 5y + 7z = 2×2 + 3×1 + 5×0 + 7×2 = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Order the five finishers by time behind Pedro.

Let Pedro's time be P.
Nico finished 11 minutes behind Pedro: N = P+11.
Olympia finished 3 minutes behind Pedro: O = P+3.
Olympia finished 2 minutes ahead of Melina: So Melina is 2 minutes behind Olympia: M = O+2 = P+5.
Olympia finished 6 minutes ahead of Luke: So Luke is 6 minutes behind Olympia: L = O+6 = P+9.

Order by increasing time:
Pedro at P+0 — 1st
Olympia at P+3 — 2nd
Melina at P+5 — 3rd
Luke at P+9 — 4th
Nico at P+11 — 5th

So fourth is Luke.

ANSWER 4: A

Problem 5:
Total elapsed time from boarding morning bus to arriving home.

From 7:30 a.m. to 4:00 p.m. = 8.5 hours = 8×60 + 30 = 510 minutes.

Time at school:
6 classes × 50 min = 300 min.
Lunch 30 min.
Additional 2 hours = 120 min.
Total school = 300 + 30 + 120 = 450 min.

Bus time (morning + afternoon) = total elapsed - school time = 510 - 450 = 60 minutes.

Tempting error: miscount 7:30 to 4:00 as 8 hours.

ANSWER 5: B

Problem 6:
Let g = number of girls. Then number of boys b = g+2.

Each boy gets b beans, so boys total = b × b = b^2.
Each girl gets g beans, so girls total = g × g = g^2.
Beans used = 400 - 6 left = 394.

So b^2 + g^2 = 394.
Substitute: (g+2)^2 + g^2 = 394
g^2 + 4g + 4 + g^2 = 394
2g^2 + 4g - 390 = 0
Divide 2: g^2 + 2g - 195 = 0.

Factor 195 = 13 × 15, differing by 2: (g+15)(g-13)=0.
Positive solution g = 13. Then b = 15.

Total students = 13+15 = 28.

Check: 15^2+13^2 =225+169=394 correct.

ANSWER 6: B

Problem 7:
The car travels 30,000 miles. At any moment exactly 4 tires are on the road carrying it.

Total tire-miles delivered = 4 tires × 30,000 miles = 120,000 tire-miles.

This wear is shared equally among 5 tires by rotation.
Per tire = 120,000 / 5 = 24,000 miles.

Tempting wrong: 30,000/5 = 6,000 ignores that 4 tires are used at once. 30,000 would mean no rotation.

ANSWER 7: C

Problem 8:
List two-digit multiples of 7:
14,21,28,35,42,49,56,63,70,77,84,91,98.

Compute digit sums:
1+4=5, 2+1=3, 2+8=10, 3+5=8, 4+2=6, 4+9=13, 5+6=11, 6+3=9, 7+0=7, 7+7=14, 8+4=12, 9+1=10, 9+8=17.

Only two have sum 10: 28 and 91.
Their sum = 28+91 = 119.

ANSWER 8: A

Problem 9:
Volume of cylinder = π r^2 h, where r = diameter/2.

Alex: diameter 6, so r=3. height 12.
V_A = π × 3^2 × 12 = π × 9 × 12 = 108π.

Felicia: diameter 12, so r=6. height 6.
V_F = π × 6^2 × 6 = π × 36 × 6 = 216π.

Ratio V_A : V_F = 108π : 216π = 108:216 = 1:2.

Note volume scales with square of radius, so doubling diameter quadruples base area, outweighing halving height.

ANSWER 9: B

Problem 10:
We need N > 2 with N ≡ 2 mod 3, mod 4, mod 5, mod 6.

So N-2 is divisible by each of 3,4,5,6.
N-2 must be a common multiple of 3,4,5,6.

LCM(3,4,5,6): 4=2^2, 6=2×3, so LCM = 2^2 × 3 × 5 = 60.

Positive multiples: 60,120,...
N=2 gives N-2=0 which works but is excluded by >2.
Next smallest is N-2 = 60, so N = 62.

62 is between 61 and 65.

Check: 62 ÷ 3 =20 rem 2, ÷4=15 rem 2, etc.

ANSWER 10: D

Problem 1:
**Restatement and Setup:**
We are given an item with an original price of $\$80$. It is discounted by $25\%$, and then a $10\%$ sales tax is applied to this discounted sale price. We need to find the final total selling price.

**Step-by-step computation:**
1. Calculate the discount:
   $$\text{Discount} = 25\% \text{ of } \$80 = 0.25 \times 80 = \$20$$
2. Calculate the sale price:
   $$\text{Sale Price} = \$80 - \$20 = \$60$$
   *(Alternatively, the sale price is $75\%$ of $\$80$: $0.75 \times 80 = \$60$.)*
3. Calculate the tax on the sale price:
   $$\text{Tax} = 10\% \text{ of } \$60 = 0.10 \times 60 = \$6$$
4. Calculate the total price:
   $$\text{Total Price} = \$60 + \$6 = \$66$$

**Ruling out wrong choices:**
- Choice A ($\$45$) results from incorrectly subtracting both $25\%$ and $10\%$ from the original price, or similar errors.
- Choice B ($\$52$) and Choice C ($\$54$) are too low.
- Choice E ($\$68$) might occur if the tax were incorrectly computed on the original $\$80$ price ($80 \times 0.10 = \$8$) and added to $\$60$.

ANSWER 1: D

---

Problem 2:
**Restatement and Setup:**
Anh drives on two segments:
- Coastal access road: distance $d_c = 10$ miles, time $t_c = 30$ minutes.
- Highway: distance $d_h = 50$ miles.
He drives three times as fast on the highway as on the coastal access road ($v_h = 3 v_c$). We need to find the total time in minutes for the entire trip.

**Step-by-step computation:**
1. Find the speed on the coastal road:
   Anh drives $10$ miles in $30$ minutes ($0.5$ hours).
   $$v_c = \frac{10\text{ miles}}{0.5\text{ hours}} = 20\text{ mph}$$
2. Find the speed on the highway:
   $$v_h = 3 \times v_c = 3 \times 20\text{ mph} = 60\text{ mph} = 1\text{ mile per minute}$$
3. Find the time spent on the highway:
   $$t_h = \frac{50\text{ miles}}{1\text{ mile per minute}} = 50\text{ minutes}$$
4. Find the total trip time:
   $$\text{Total time} = t_c + t_h = 30\text{ minutes} + 50\text{ minutes} = 80\text{ minutes}$$

**Ruling out wrong choices:**
- Choice A ($50$) is just the time on the highway.
- Choices B ($70$), D ($90$), and E ($100$) arise from miscalculating the highway speed ratio or mixing up minutes and hours.

ANSWER 2: C

---

Problem 3:
**Restatement and Setup:**
Given whole numbers (non-negative integers) $w, x, y, z$ such that:
$$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$$
We need to find the value of $2w + 3x + 5y + 7z$.

**Step-by-step computation:**
1. Find the prime factorization of $588$:
   $$588 = 2 \times 294 = 2 \times 2 \times 147 = 4 \times 3 \times 49 = 2^2 \times 3^1 \times 7^2$$
2. Matching the prime factors with $2^w \cdot 3^x \cdot 5^y \cdot 7^z$:
   - $w = 2$
   - $x = 1$
   - $y = 0$ (since $5$ is not a factor of $588$)
   - $z = 2$
3. Compute the required expression $2w + 3x + 5y + 7z$:
   $$2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21$$

**Ruling out wrong choices:**
- If one mistakenly sets $y=1$, the sum would be $21 + 5 = 26$ (close to 25 or 27).
- Choices B, C, D, and E do not match the evaluated sum of $21$.

ANSWER 3: A

---

Problem 4:
**Restatement and Setup:**
Five runners (Luke, Melina, Nico, Olympia, Pedro) finished a race. We are given relative finish times:
- Nico finished $11$ minutes behind Pedro.
- Olympia finished $3$ minutes behind Pedro.
- Olympia finished $2$ minutes ahead of Melina.
- Olympia finished $6$ minutes ahead of Luke.
We need to determine who finished in fourth place.

**Step-by-step computation:**
Let Pedro's finish time be $t=0$ minutes (the earliest/winner).
- Pedro finishes at $0$ minutes.
- Olympia finished $3$ minutes behind Pedro $\implies$ Olympia finishes at $0 + 3 = 3$ minutes.
- Olympia finished $2$ minutes ahead of Melina $\implies$ Melina finishes $2$ minutes after Olympia $\implies 3 + 2 = 5$ minutes.
- Olympia finished $6$ minutes ahead of Luke $\implies$ Luke finishes $6$ minutes after Olympia $\implies 3 + 6 = 9$ minutes.
- Nico finished $11$ minutes behind Pedro $\implies$ Nico finishes at $0 + 11 = 11$ minutes.

Listing the runners in order of finish:
1. Pedro ($0$ min)
2. Olympia ($3$ min)
3. Melina ($5$ min)
4. Luke ($9$ min)
5. Nico ($11$ min)

Thus, Luke finished fourth.

**Ruling out wrong choices:**
- Pedro was 1st, Olympia was 2nd, Melina was 3rd, and Nico was 5th (last).

ANSWER 4: A

---

Problem 5:
**Restatement and Setup:**
Walter boards the bus at 7:30 a.m. and arrives home from the bus at 4:00 p.m.
During the day, he spends time on the bus (both ways) and at school.
School activities consist of:
- $6$ classes of $50$ minutes each
- $30$ minutes for lunch
- $2$ hours of additional time
We need to find the total time Walter spent on the bus.

**Step-by-step computation:**
1. Calculate total elapsed time between leaving home and returning:
   - From 7:30 a.m. to 12:00 p.m. is $4$ hours and $30$ minutes ($4.5$ hours).
   - From 12:00 p.m. to 4:00 p.m. is $4$ hours.
   - Total elapsed time = $8$ hours and $30$ minutes $= 8 \times 60 + 30 = 510$ minutes.
2. Calculate total time spent on school activities:
   - Classes: $6 \times 50\text{ minutes} = 300\text{ minutes}$
   - Lunch: $30\text{ minutes}$
   - Additional time: $2\text{ hours} = 120\text{ minutes}$
   - Total school time $= 300 + 30 + 120 = 450\text{ minutes}$.
3. Calculate time spent on the bus:
   $$\text{Bus time} = \text{Total time} - \text{School time} = 510 - 450 = 60\text{ minutes}$$

**Ruling out wrong choices:**
- 30 minutes assumes only a one-way trip or forgets additional school time.
- 75, 90, 120 minutes result from miscounting the elapsed time or forgetting the 30-minute lunch.

ANSWER 5: B

---

Problem 6:
**Restatement and Setup:**
Let $b$ be the number of boys and $g$ be the number of girls.
- Each boy receives $b$ jelly beans (total $b^2$).
- Each girl receives $g$ jelly beans (total $g^2$).
- Total jelly beans distributed: $400 - 6 = 394$.
- There are two more boys than girls: $b = g + 2$.
We need to find the total number of students in the class, which is $b + g$.

**Step-by-step com

# Problem 1:
**What is asked:** Find the final price after a 25% discount on $80, followed by 10% tax on the *sale price*.

**Step 1 — Compute the sale price (25% off):**
Discount = 25% of 80 = 0.25 × 80 = $20.
Sale price = 80 − 20 = **$60**.

**Step 2 — Add 10% tax to the sale price:**
Tax = 10% of 60 = $6.
Total = 60 + 6 = **$66**.

**Rule out wrong answers:**
- $52 (B) comes from wrongly taking 10% off the *original* price (60 − 8), but tax is added, not discounted.
- $54 (C) comes from adding discounts: 80 × 0.75 × 0.9 — but tax increases the price.
- The correct sequence is 80 → 60 → 66.

ANSWER 1: D

# Problem 2:
**What is asked:** Find the total driving time in minutes.

**Step 1 — Coastal road speed:**
Anh drove 10 miles in 30 minutes (= 0.5 hour).
Speed = 10 ÷ 0.5 = **20 mph**.

**Step 2 — Highway speed:**
He drove 3 times as fast: 3 × 20 = **60 mph**.

**Step 3 — Highway time:**
Time = distance ÷ speed = 50 ÷ 60 = 5/6 hour = **50 minutes**.

**Step 4 — Total time:**
30 + 50 = **80 minutes**.

**Rule out wrong answers:** 
- 90 (D) would result from wrongly assuming equal speeds (30 min + 60 min); but the highway was faster, so its time is *less* than 60 minutes. Only 80 fits.

ANSWER 2: C

# Problem 3:
**What is asked:** Given 2^w · 3^x · 5^y · 7^z = 588, find 2w + 3x + 5y + 7z.

**Step 1 — Factor 588:**
588 = 4 × 147 = 4 × 3 × 49 = 2² × 3 × 7².

**Step 2 — Read off exponents:**
588 = 2² · 3¹ · 5⁰ · 7², so w = 2, x = 1, y = 0, z = 2.

**Step 3 — Compute the expression:**
2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = **21**.

**Rule out wrong answers:** 35 (D) or 56 (E) would come from confusing the coefficients 2,3,5,7 with the bases; note 588 is not divisible by 5, forcing y = 0.

ANSWER 3: A

# Problem 4:
**What is asked:** Determine which runner finished fourth (in order of finishing time).

**Step 1 — Set Pedro's time as P (a reference).** Then:
- Nico finished 11 minutes behind Pedro: Nico = P + 11.
- Olympia finished 3 minutes behind Pedro: Olympia = P + 3.
- Olympia finished 2 minutes ahead of Melina, so Melina = P + 3 + 2 = P + 5.
- Olympia finished 6 minutes ahead of Luke, so Luke = P + 3 + 6 = P + 9.

**Step 2 — Order from fastest to slowest:**
Pedro (P), Olympia (P+3), Melina (P+5), Luke (P+9), Nico (P+11).

**Step 3 — Fourth place** is **Luke**.

**Rule out wrong answers:** Nico (C) is last (5th), Pedro (E) is first; a common error is swapping Luke and Melina — but Luke = P+9 is behind Melina = P+5.

ANSWER 4: A

# Problem 5:
**What is asked:** Total time Walter spent on the bus (both ways).

**Step 1 — Total elapsed time from 7:30 a.m. to 4:00 p.m.:**
That is 8 hours 30 minutes = 510 minutes.

**Step 2 — Time accounted for at school:**
- Classes: 6 × 50 = 300 minutes
- Lunch: 30 minutes
- Additional time: 2 hours = 120 minutes
- Total school time: 300 + 30 + 120 = 450 minutes.

**Step 3 — Bus time:**
510 − 450 = **60 minutes**.

**Rule out wrong answers:** 90 (D) would ignore lunch or additional time; 30 (A) would count only one bus ride — but the round trip must be 60 minutes total.

ANSWER 5: B

# Problem 6:
**What is asked:** Find the total number of students.

**Step 1 — Set up equations.** Let g = number of girls, b = number of boys.
Each boy gets b beans (b boys), each girl gets g beans (g girls):
b² + g² = 400 − 6 = 394, and b = g + 2.

**Step 2 — Substitute and solve:**
(g+2)² + g² = 394
g² + 4g + 4 + g² = 394
2g² + 4g − 390 = 0
g² + 2g − 195 = 0
(g + 15)(g − 13) = 0 → g = 13 (reject −15).

**Step 3 — Find the total:**
Boys: b = 13 + 2 = 15.
Total students = 15 + 13 = **28**.

**Check:** 15² + 13² = 225 + 169 = 394 = 400 − 6. ✓

ANSWER 6: B

# Problem 7:
**What is asked:** Miles each of the 5 tires was used over 30,000 miles.

**Step 1 — Total tire-miles used:**
At any moment the car uses 4 tires. Over 30,000 miles, total tire-miles = 4 × 30,000 = 120,000.

**Step 2 — Split equally among 5 tires:**
120,000 ÷ 5 = **24,000 miles** each.

**Rule out wrong answers:** 30,000 (D) is wrong because no single tire can be used the whole time when 5 tires share the work; 6,000 (A) and 7,500 (B) wrongly divide by the wrong quantity (e.g., 30,000 ÷ 5 or ÷ 4).

ANSWER 7: C

# Problem 8:
**What is asked:** Sum of the two two-digit multiples of 7 whose digits sum to 10.

**Step 1 — List two-digit multiples of 7 and digit sums:**
14 (5), 21 (3), **28 (10)** ✓, 35 (8), 42 (6), 49 (13), 56 (11), 63 (9), 70 (7), 77 (14), 84 (12), **91 (10)** ✓, 98 (17).

**Step 2 — Identify the two qualifying numbers:** 28 and 91.

**Step 3 — Sum:** 28 + 91 = **119**.

**Rule out wrong answers:** 126 (B) is 63 + 63 — impossible since numbers are distinct; 140 (D) might come from 28 + 112 (but 112 is three digits).

ANSWER 8: A

# Problem 9:
**What is asked:** Ratio of Alex's can volume to Felicia's can volume.

**Step 1 — Volume formula:** V = πr²h.

**Step 2 — Alex's can:** diameter 6 → radius 3; height 12.
V_A = π(3)²(12) = 108π.

**Step 3 — Felicia's can:** diameter 12 → radius 6; height 6.
V_F = π(6)²(6) = 216π.

**Step 4 — Ratio:**
V_A : V_F = 108π : 216π = **1 : 2**.

**Rule out wrong answers:** 1:1 (C) is the tempting "swapped dimensions" intuition, but volume depends on the *square* of the radius, so doubling the radius quadruples that factor while halving the height only halves — net effect is Felicia's can is twice as big.

ANSWER 9: B

# Problem 10:
**What is asked:** Find the smallest number > 2 that leaves remainder 2 when divided by 3, 4, 5, and 6, and identify the interval containing it.

**Step 1 — Interpret the condition:**
N − 2 must be divisible by 3, 4, 5, and 6, i.e., N − 2 is a multiple of LCM(3, 4, 5, 6).

**Step 2 — Compute the LCM:**
LCM(3,4,5,6) = 60 (since 60 is divisible by 3, 4, 5, and 6).

**Step 3 — Candidates:**
N = 60k + 2 for whole numbers k: N = 2, 62, 122, ...
The smallest *greater than 2* is **62**.

**Step 4 — Locate 62:** It lies between **61 and 65**.

**Rule out wrong answers:** 42 + 2 = 44 is a common error (using 42, the LCM of only 

Problem 1:  
The dress is originally priced at \$80. A 25% discount means the sale price is

\[
80(1-0.25)=80(0.75)=60.
\]

Then 10% tax is added to the sale price:

\[
60(1+0.10)=60(1.10)=66.
\]

So the total selling price is \$66.

ANSWER 1: D

Problem 2:  
Anh drove 10 miles on the coastal road in 30 minutes, so his coastal-road speed was

\[
\frac{10\text{ miles}}{30\text{ minutes}}.
\]

He drove three times as fast on the highway. Therefore, for the same distance, the highway time would be one-third as long.

For 50 miles on the highway, if he had driven at coastal-road speed, it would take

\[
30 \times \frac{50}{10}=150
\]

minutes. But because he drove three times as fast on the highway, the actual highway time is

\[
\frac{150}{3}=50
\]

minutes.

The coastal road took 30 minutes, so the total trip time was

\[
30+50=80
\]

minutes.

ANSWER 2: C

Problem 3:  
The equation should be read as

\[
2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.
\]

Factor 588:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49=2^2\cdot 3^1\cdot 7^2.
\]

There is no factor of 5, so \(y=0\). Comparing prime exponents gives

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2.
\]

Now compute

\[
2w+3x+5y+7z=2(2)+3(1)+5(0)+7(2).
\]

\[
=4+3+0+14=21.
\]

ANSWER 3: A

Problem 4:  
Let Pedro’s finishing time be \(P\). In a race, a smaller time means finishing earlier.

- Nico finished 11 minutes behind Pedro, so Nico’s time is

\[
P+11.
\]

- Olympia finished 3 minutes behind Pedro, so Olympia’s time is

\[
P+3.
\]

- Olympia finished 2 minutes ahead of Melina, so Melina’s time is

\[
(P+3)+2=P+5.
\]

- Olympia finished 6 minutes ahead of Luke, so Luke’s time is

\[
(P+3)+6=P+9.
\]

Now order the runners from first to last by increasing time:

\[
P,\quad P+3,\quad P+5,\quad P+9,\quad P+11.
\]

That is:

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

So the fourth runner was Luke.

ANSWER 4: A

Problem 5:  
From 7:30 a.m. to 4:00 p.m. is 8 hours and 30 minutes:

\[
8(60)+30=480+30=510
\]

minutes.

Walter’s time in school not spent on the bus is:

- 6 classes of 50 minutes each:

\[
6(50)=300
\]

minutes;

- lunch: 30 minutes;

- additional time: 2 hours = 120 minutes.

Total non-bus time:

\[
300+30+120=450
\]

minutes.

So his total bus time was

\[
510-450=60
\]

minutes.

ANSWER 5: B

Problem 6:  
Let the number of girls be \(g\). Since there were two more boys than girls, the number of boys is

\[
g+2.
\]

Mrs. Awesome gave each boy as many jelly beans as there were boys, so the boys received

\[
(g+2)^2
\]

jelly beans total. Each girl received as many as there were girls, so the girls received

\[
g^2
\]

jelly beans total.

She had 400 jelly beans and 6 left, so she gave away

\[
400-6=394.
\]

Thus,

\[
(g+2)^2+g^2=394.
\]

Expand:

\[
g^2+4g+4+g^2=394
\]

\[
2g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0.
\]

Factor:

\[
(g+15)(g-13)=0.
\]

Since \(g\) must be positive,

\[
g=13.
\]

Then the number of boys is

\[
13+2=15.
\]

The total number of students is

\[
13+15=28.
\]

ANSWER 6: B

Problem 7:  
During the first 30,000 miles, the car was always using 4 tires at once. Therefore, the total “tire-miles” used is

\[
4 \times 30{,}000=120{,}000.
\]

There are 5 tires total, and they were all used the same number of miles. So each tire was used for

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

ANSWER 7: C

Problem 8:  
We need two-digit multiples of 7 whose digits sum to 10.

The two-digit multiples of 7 are:

\[
14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
\]

Check digit sums:

\[
14: 1+4=5
\]
\[
21: 2+1=3
\]
\[
28: 2+8=10
\]
\[
35: 3+5=8
\]
\[
42: 4+2=6
\]
\[
49: 4+9=13
\]
\[
56: 5+6=11
\]
\[
63: 6+3=9
\]
\[
70: 7+0=7
\]
\[
77: 7+7=14
\]
\[
84: 8+4=12
\]
\[
91: 9+1=10
\]
\[
98: 9+8=17
\]

The only two are 28 and 91. Their sum is

\[
28+91=119.
\]

ANSWER 8: A

Problem 9:  
The volume of a cylinder is

\[
V=\pi r^2h.
\]

Alex’s cans have diameter 6 cm, so radius 3 cm, and height 12 cm:

\[
V_A=\pi(3^2)(12)=\pi(9)(12)=108\pi.
\]

Felicia’s cans have diameter 12 cm, so radius 6 cm, and height 6 cm:

\[
V_F=\pi(6^2)(6)=\pi(36)(6)=216\pi.
\]

The ratio of Alex’s can volume to Felicia’s can volume is

\[
108\pi:216\pi=108:216=1:2.
\]

ANSWER 9: B

Problem 10:  
Let the number be \(N\). It leaves a remainder of 2 when divided by 3, 4, 5, and 6, so

\[
N-2
\]

is divisible by 3, 4, 5, and 6. Therefore, \(N-2\) must be a multiple of the least common multiple of 3, 4, 5, and 6.

Find the LCM:

\[
\operatorname{lcm}(3,4,5,6)=60.
\]

So

\[
N-2=60k
\]

for some positive integer \(k\), because we need \(N>2\).

The smallest possible value is when \(k=1\):

\[
N-2=60
\]

\[
N=62.
\]

This lies between 61 and 65.

ANSWER 10: D

Problem 1:
Sale price: 25% off means you pay 75%: 80 × 0.75 = 60 dollars.
Add 10% tax: 60 × 1.10 = 66 dollars.
(Choice C, 54, would come from subtracting 10%; choice D is correct. Note taking 25% off then adding 10% doesn't return to 80.)

ANSWER 1: D

Problem 2:
Coastal road: 10 miles in 30 minutes, so 3 minutes per mile.
Highway speed is 3 times as fast, so 1 minute per mile.
Highway time: 50 miles × 1 minute/mile = 50 minutes.
Total: 50 + 30 = 80 minutes.
(Choice A forgets the coastal 30 minutes; choice D would come from using 3 min/mile on the highway incorrectly.)

ANSWER 2: C

Problem 3:
The equation is 2^w · 3^x · 5^y · 7^z = 588.
Factor 588 = 4 · 147 = 2² · 3 · 7².
So w = 2, x = 1, y = 0, z = 2.
Then 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21.

ANSWER 3: A

Problem 4:
Let Pedro's time be 0 (he's a reference).
- Nico = +11
- Olympia = +3 (3 min behind Pedro)
- Melina = Olympia + 2 = +5
- Luke = Olympia + 6 = +9
Order from first: Pedro (0), Olympia (3), Melina (5), Luke (9), Nico (11).
Fourth place is Luke.

ANSWER 4: A

Problem 5:
Total time away: 7:30 a.m. to 4:00 p.m. = 8 hours 30 minutes = 510 minutes.
Time at school: classes 6 × 50 = 300, lunch 30, extra 2 hours = 120; total 450 minutes.
Bus time = 510 − 450 = 60 minutes (both trips combined).

ANSWER 5: B

Problem 6:
Let g = number of girls, b = g + 2 boys.
Jelly beans used: b² + g² = 400 − 6 = 394.
(g + 2)² + g² = 394 → 2g² + 4g + 4 = 394 → g² + 2g − 195 = 0 → (g + 15)(g − 13) = 0 → g = 13.
So girls = 13, boys = 15; check 169 + 225 = 394 ✓.
Total students = 28.

ANSWER 6: B

Problem 7:
Four tires are on the road at any time, so total tire-miles = 4 × 30,000 = 120,000.
Shared equally among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
(30,000 would be if no rotation; 6,000 is the rest time per tire, not use.)

ANSWER 7: C

Problem 8:
Two-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Digit sums: 5, 3, 10, 8, 6, 13, 11, 9, 7, 14, 12, 10, 17.
Digit sum 10 occurs for 28 and 91.
Sum: 28 + 91 = 119.

ANSWER 8: A

Problem 9:
Alex: diameter 6 → radius 3, height 12: V = π(3²)(12) = 108π.
Felicia: diameter 12 → radius 6, height 6: V = π(6²)(6) = 216π.
Ratio = 108π : 216π = 1 : 2.
(Doubling the radius quadruples volume, while halving height only halves it, so Felicia's is twice as big — not equal.)

ANSWER 9: B

Problem 10:
A number leaving remainder 2 when divided by 3, 4, 5, and 6 must be 2 more than a common multiple of 3, 4, 5, 6.
LCM(3, 4, 5, 6) = 60, so the smallest such number greater than 2 is 60 + 2 = 62.
62 lies between 61 and 65.

ANSWER 10: D

Problem 1:

The dress was originally priced at \$80.

A 25% discount means the sale price is \(75\%\) of the original price:

\[
80 \times 0.75=60
\]

Then 10% tax is added:

\[
10\% \text{ of } 60 = 6
\]

So the total price is

\[
60+6=66
\]

A common mistake is to subtract 25% and add 10% of the original \$80, but the tax is added to the reduced sale price, not the original price.

\[
\boxed{66}
\]

ANSWER 1: D

Problem 2:

Anh drove 10 miles on the coastal road in 30 minutes.

Since 30 minutes is \(\frac12\) hour, his coastal-road speed was

\[
\frac{10\text{ miles}}{\frac12\text{ hour}}=20\text{ miles per hour}
\]

He drove three times as fast on the highway:

\[
3(20)=60\text{ miles per hour}
\]

The highway trip was 50 miles, so the highway driving time was

\[
\frac{50}{60}\text{ hour}=\frac56\text{ hour}
\]

Convert \(\frac56\) hour to minutes:

\[
\frac56\cdot 60=50\text{ minutes}
\]

His total driving time was

\[
30+50=80\text{ minutes}
\]

The answer is not 50 minutes, since that is only the highway time.

\[
\boxed{80}
\]

ANSWER 2: C

Problem 3:

We are told that

\[
2^w\cdot 3^x\cdot 5^y\cdot 7^z=588
\]

Factor 588 into primes:

\[
588=2\cdot 294=2^2\cdot 147=2^2\cdot 3\cdot 49
\]

Thus,

\[
588=2^2\cdot 3^1\cdot 5^0\cdot 7^2
\]

So

\[
w=2,\qquad x=1,\qquad y=0,\qquad z=2
\]

Now calculate:

\[
2w+3x+5y+7z
\]

\[
=2(2)+3(1)+5(0)+7(2)
\]

\[
=4+3+0+14=21
\]

The factor \(5^0\) is important: since 588 has no factor of 5, \(y=0\), not 1.

\[
\boxed{21}
\]

ANSWER 3: A

Problem 4:

We compare all runners’ finishing times relative to Pedro.

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

Let Pedro finish at time 0.

Then:

\[
\begin{array}{c|c}
\text{Runner} & \text{Minutes behind Pedro} \\
\hline
\text{Pedro} & 0\\
\text{Olympia} & 3\\
\text{Melina} & 3+2=5\\
\text{Luke} & 3+6=9\\
\text{Nico} & 11
\end{array}
\]

Therefore, the order is:

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

So Luke finished fourth.

\[
\boxed{\text{Luke}}
\]

ANSWER 4: A

Problem 5:

Walter is at school from 7:30 a.m. until 4:00 p.m.

That is:

\[
8\text{ hours }30\text{ minutes}=510\text{ minutes}
\]

His school-day activities take:

- 6 classes of 50 minutes each:

\[
6\cdot 50=300\text{ minutes}
\]

- Lunch:

\[
30\text{ minutes}
\]

- Additional school time:

\[
2\text{ hours}=120\text{ minutes}
\]

Total time not spent on the bus:

\[
300+30+120=450\text{ minutes}
\]

Therefore, bus time is

\[
510-450=60\text{ minutes}
\]

This includes both the ride to school and the ride home.

\[
\boxed{60}
\]

ANSWER 5: B

Problem 6:

Let the number of girls be \(g\).

There are two more boys than girls, so the number of boys is

\[
g+2
\]

Each boy receives as many jelly beans as there are boys, so the boys receive

\[
(g+2)(g+2)=(g+2)^2
\]

jelly beans altogether.

Each girl receives \(g\) jelly beans, so the girls receive

\[
g^2
\]

jelly beans altogether.

Mrs. Awesome gave away

\[
400-6=394
\]

jelly beans. Therefore,

\[
g^2+(g+2)^2=394
\]

Expand:

\[
g^2+g^2+4g+4=394
\]

\[
2g^2+4g-390=0
\]

Divide by 2:

\[
g^2+2g-195=0
\]

Factor:

\[
(g+15)(g-13)=0
\]

So

\[
g=13
\]

since a number of girls cannot be negative.

Thus boys:

\[
13+2=15
\]

Total students:

\[
13+15=28
\]

\[
\boxed{28}
\]

ANSWER 6: B

Problem 7:

During the first 30,000 miles, the car always uses 4 tires on the road.

So the total number of tire-miles used is

\[
4\cdot 30{,}000=120{,}000
\]

There are 5 tires total, and they were rotated so each tire was used equally.

Therefore, each tire was used

\[
\frac{120{,}000}{5}=24{,}000
\]

miles.

It is not 30,000 miles because only four of the five tires are being driven on at any one time.

\[
\boxed{24{,}000}
\]

ANSWER 7: C

Problem 8:

We need two-digit multiples of 7 whose digits add to 10.

The two-digit multiples of 7 are:

\[
14,21,28,35,42,49,56,63,70,77,84,91,98
\]

Check digit sums:

\[
28:\quad 2+8=10
\]

\[
91:\quad 9+1=10
\]

These are the only two.

Their sum is

\[
28+91=119
\]

\[
\boxed{119}
\]

ANSWER 8: A

Problem 9:

The volume of a cylinder is

\[
V=\pi r^2h
\]

For Alex’s can:

- Diameter \(=6\), so radius \(=3\)
- Height \(=12\)

\[
V_A=\pi(3)^2(12)=108\pi
\]

For Felicia’s can:

- Diameter \(=12\), so radius \(=6\)
- Height \(=6\)

\[
V_F=\pi(6)^2(6)=216\pi
\]

The ratio of Alex’s volume to Felicia’s volume is

\[
108\pi:216\pi
\]

Divide both parts by \(108\pi\):

\[
1:2
\]

Although Alex’s can is twice as tall, Felicia’s radius is twice as large, and radius is squared in the volume formula.

\[
\boxed{1:2}
\]

ANSWER 9: B

Problem 10:

We seek the smallest number greater than 2 that leaves remainder 2 when divided by 3, 4, 5, and 6.

If the number is \(n\), then \(n-2\) must be divisible by all of

\[
3,\ 4,\ 5,\ 6
\]

The least common multiple is

\[
\operatorname{lcm}(3,4,5,6)=60
\]

So the smallest possible value of \(n-2\) is 60:

\[
n-2=60
\]

\[
n=62
\]

The number 62 lies between 61 and 65.

Indeed:

\[
62\div 3,\ 62\div 4,\ 62\div 5,\ 62\div 6
\]

all leave remainder 2.

\[
\boxed{61\text{ and }65}
\]

ANSWER 10: D