🔬 Model Lab

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A

Problem 1:
We are asked: start with 200 gallons, lose 0.5 gallons per day for 30 days, how much is left?

Set up: Total loss = rate × time.
Rate = 0.5 gallons/day, time = 30 days.
Total loss = 0.5 × 30 = 15 gallons.

Remaining = starting - loss = 200 - 15 = 185 gallons.

Check wrong choices: D 198.5 is loss for only 3 days. E 199.85 confuses 0.5 per day with 0.005? A 140 and B 170 are far too much loss.
ANSWER 1: C

Problem 2:
We are asked to apply the rule for a 2x2 determinant. The symbol

3 1
4 2

with top row a=3, c=1 and bottom row b=4, d=2 is defined to be a·d − b·c.

Set up: a·d − b·c = 3·2 − 4·1.
Compute: 3·2 = 6, 4·1 = 4, so 6 − 4 = 2.

Check: Reversing the subtraction gives -2, choice A, which is wrong because order is a·d minus b·c, not the reverse. The result is 2.
ANSWER 2: E

Problem 3:
We are asked for the sum of the (distinct) prime factors of 2010.

Factor 2010: 2010 = 201 × 10 = 3 × 67 × 2 × 5.
Check primality: 2,3,5 are prime, 67 is prime. 2010 is squarefree, so these are all.

Sum = 2 + 3 + 5 + 67 = 77.
Compute: 2+3+5=10, 10+67=77.

Wrong choices: A 67 omits the small factors. B 75 is arithmetic error. D 201 and E 210 are factors, not sum of prime factors.
ANSWER 3: C

Problem 4:
We are asked what fraction of the final mixture poured together is orange juice.

Each pitcher holds 600 mL. Orange juice amounts:
Pitcher 1: 600 × 1/3 = 200 mL.
Pitcher 2: 600 × 2/5 = 1200/5 = 240 mL.
Total orange juice = 200 + 240 = 440 mL.

Total mixture after filling with water and pouring together = 600 + 600 = 1200 mL.
Fraction juice = 440 / 1200 = 44 / 120 = 11 / 30.

This is also the average of 1/3 and 2/5: (1/3+2/5)/2 = (11/15)/2 = 11/30.
Wrong: 1/8, 3/16 are too small; 11/15 forgets to divide by 2 pitchers; 11/19 miscomputes total.
ANSWER 4: C

Problem 5:
We are asked to compute each person's final number from starting number 10 with different order of operations, and compare.

Jose: subtract 1, then double, then add 2:
10-1=9; 9×2=18; 18+2=20.

Thuy: double, then subtract 1, then add 2:
10×2=20; 20-1=19; 19+2=21.

Kareem: subtract 1, then add 2, then double:
10-1=9; 9+2=11; 11×2=22.

Compare: 20, 21, 22. Largest is Kareem with 22.

So A, B, D, E are wrong; only Kareem is largest.
ANSWER 5: C

Problem 6:
We are asked what fraction of a 12-slice pizza Peter ate, if he ate 1 whole slice plus half of another slice.

Peter's slices = 1 + 1/2 = 3/2 slices.
Total slices = 12.
Fraction = (3/2) / 12 = 3 / 24 = 1/8.

Wrong: A 1/24 is only half a slice out of 12. B 1/12 is one slice out of 12, forgetting the shared half. D 1/6 = 2/12 forgets sharing. E 1/4 = 3/12 counts the whole shared slice.
ANSWER 6: C

Problem 7:
We are asked for the minimum third-test score that still allows an average of 95 over four 100-point tests.

Goal total = 4 × 95 = 380 points.
First two total = 97 + 91 = 188.
Still needed from tests 3 and 4 = 380 - 188 = 192.

Let third score = x, fourth ≤ 100. To still be possible we need x + 100 ≥ 192, so x ≥ 92.
If x = 92, she can get 100 on test 4 to reach exactly 380. If x < 92, even 100 is not enough.

So lowest possible x that keeps the goal alive is 92.
Wrong: A 90 would require 102 on last test, impossible. C,D,E are possible but not lowest.
ANSWER 7: B

Problem 8:
We are asked to find the correct sunset time given sunrise 6:57 AM and daylight length 10 hours 24 minutes.

Sunset = sunrise + daylight length.
6:57 AM + 10 hours = 4:57 PM.
4:57 PM + 24 minutes = 5:21 PM.

Check: 8:15 PM listed would imply over 13 hours daylight, inconsistent with 10h24m, so it is wrong. 5:21 PM fits.
ANSWER 8: B

Problem 9:
We are asked in how many weeks of paper-route earnings plus birthday money Chandler can afford a $500 bike.

Birthday money = 50 + 35 + 15 = 100 dollars.
Still needed = 500 - 100 = 400 dollars.
Pay = $16 per week. Weeks needed = 400 / 16 = 25.
Check: 16 × 25 = 400 exactly.

Wrong: A 24 gives only 16×24+100 = 484 < 500. C,D,E are more than needed.
ANSWER 9: B

Problem 10:
We are asked how many of 39 students have both a dog and a cat, given 20 have a dog, 26 have a cat, and everyone has at least one.

Use inclusion-exclusion: Total with at least one = Dog-only + Cat-only + Both = Dog + Cat − Both, because Both was counted twice.
So 39 = 20 + 26 − Both.
Thus Both = 20 + 26 − 39 = 46 − 39 = 7.

Wrong: B 13, C 19 miscount; D 39 would mean all have both; E 46 is Dog+Cat without subtracting overlap.
ANSWER 10: A

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

---

### Problem 1:
**Restatement:** A pool starts with $200$ gallons of water and loses $0.5$ gallons per day due to evaporation. We need to find the amount of water remaining after $30$ days.

**Step-by-step solution:**
1. Calculate the total volume of water evaporated over $30$ days:
   $$\text{Water evaporated} = 30 \text{ days} \times 0.5 \text{ gallons/day} = 15 \text{ gallons}$$
2. Subtract this amount from the starting volume:
   $$\text{Water remaining} = 200 - 15 = 185 \text{ gallons}$$

Therefore, after $30$ days, there will be $185$ gallons left.

ANSWER 1: C

---

### Problem 2:
**Restatement:** We are given the operation representing a $2 \times 2$ determinant: 
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$$
We need to evaluate this expression for the numbers $3, 1, 4, 2$, where $a = 3$, $b = 1$, $c = 4$, and $d = 2$.

**Step-by-step solution:**
1. Substitute the values into the formula $a \cdot d - b \cdot c$:
   $$3 \cdot 2 - 1 \cdot 4$$
2. Compute the products:
   $$3 \cdot 2 = 6$$
   $$1 \cdot 4 = 4$$
3. Subtract the results:
   $$6 - 4 = 2$$

ANSWER 2: E

---

### Problem 3:
**Restatement:** Find the sum of the prime factors of $2010$.

**Step-by-step solution:**
1. Find the prime factorization of $2010$:
   - $2010$ is even: $2010 = 2 \times 1005$
   - $1005$ ends in $5$: $1005 = 5 \times 201$
   - The sum of the digits of $201$ is $2 + 0 + 1 = 3$, so it is divisible by $3$: $201 = 3 \times 67$
   - $67$ is a prime number.
2. The distinct prime factors of $2010$ are $2, 3, 5,$ and $67$.
3. Sum the prime factors:
   $$2 + 3 + 5 + 67 = 77$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Both are filled to capacity with water and poured into a single container. What fraction of the final mixture is orange juice?

**Step-by-step solution:**
1. Find the amount of orange juice in the first pitcher:
   $$\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$$
2. Find the amount of orange juice in the second pitcher:
   $$\frac{2}{5} \times 600\text{ mL} = 240\text{ mL}$$
3. Find the total volume of orange juice:
   $$200\text{ mL} + 240\text{ mL} = 440\text{ mL}$$
4. Find the total volume of the combined mixture from both completely filled pitchers:
   $$600\text{ mL} + 600\text{ mL} = 1200\text{ mL}$$
5. Calculate the fraction of the mixture that is orange juice:
   $$\frac{440}{1200} = \frac{44}{120} = \frac{11}{30}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** Jose, Thuy, and Kareem perform different arithmetic operations starting with the number $10$. We must determine who ends with the largest result.

**Step-by-step solution:**
1. **Jose:**
   - Subtract $1$: $10 - 1 = 9$
   - Double the answer: $9 \times 2 = 18$
   - Add $2$: $18 + 2 = 20$
2. **Thuy:**
   - Double $10$: $10 \times 2 = 20$
   - Subtract $1$: $20 - 1 = 19$
   - Add $2$: $19 + 2 = 21$
3. **Kareem:**
   - Subtract $1$: $10 - 1 = 9$
   - Add $2$: $9 + 2 = 11$
   - Double the result: $11 \times 2 = 22$

Comparing the three final numbers ($20$, $21$, and $22$), Kareem gets the largest final answer ($22$).

ANSWER 5: C

---

### Problem 6:
**Restatement:** A pizza has $12$ equal slices. Peter ate one whole slice and half of another slice. What fraction of the entire pizza did Peter eat?

**Step-by-step solution:**
1. Determine the total number of slices Peter ate:
   $$\text{Slices Peter ate} = 1 + \frac{1}{2} = 1.5\text{ slices} = \frac{3}{2}\text{ slices}$$
2. Find the fraction of the $12$-slice pizza that this represents:
   $$\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$$

ANSWER 6: C

---

### Problem 7:
**Restatement:** Isabella takes four $100$-point tests and aims for an average of $95$. Her first two test scores are $97$ and $91$. What is the lowest possible score she could have achieved on the third test such that she could still reach her goal on the fourth test?

**Step-by-step solution:**
1. Calculate the total number of points needed across all four tests to achieve an average of $95$:
   $$\text{Total points needed} = 4 \times 95 = 380$$
2. Sum her scores from the first two tests:
   $$97 + 91 = 188$$
3. Find the total points needed from the third and fourth tests combined:
   $$380 - 188 = 192$$
4. To minimize the score on the third test, Isabella must maximize the score on the fourth test. The maximum possible score on any test is $100$:
   $$\text{Lowest score on test 3} = 192 - 100 = 92$$

ANSWER 7: B

---

### Problem 8:
**Restatement:** Given that sunrise was at $6:57\text{ AM}$ and the daylight lasted $10\text{ hours}$ and $24\text{ minutes}$, find the correct sunset time.

**Step-by-step solution:**
1. Add $10\text{ hours}$ to the sunrise time of $6:57\text{ AM}$:
   $$6:57\text{ AM} + 10\text{ hours} = 4:57\text{ PM}$$
2. Add the remaining $24\text{ minutes}$ to $4:57\text{ PM}$:
   $$4:57\text{ PM} + 24\text{ minutes} = 5:21\text{ PM}$$

Thus, the sun actually set at $5:21\text{ PM}$.

ANSWER 8: B

---

### Problem 9:
**Restatement:** Chandler wants to buy a $\$500$ bike. He receives $\$50$, $\$35$, and $\$15$ for his birthday and earns $\$16$ per week from a paper route. In how many weeks will he have enough money to buy the bike?

**Step-by-step solution:**
1. Calculate the total birthday money received:
   $$50 + 35 + 15 = \$100$$
2. Determine how much more money he needs to save:
   $$500 - 100 = \$400$$
3. Divide the remaining amount by his weekly earnings:
   $$\text{Number of weeks} = \frac{400}{16} = 25\text{ weeks}$$

ANSWER 9: B

---

### Problem 10:
**Restatement:** In a class of $39$ students, each student has a dog, a cat, or both. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

**Step-by-step solution:**
1. Apply the Principle of Inclusion-Exclusion for sets:
   $$|D \cup C| = |D| + |C| - |D \cap C|$$
2. Plug in the give

# Problem 1:
**Restating:** A pool starts with 200 gallons. Water evaporates at 0.5 gallons per day for 30 days. Find the remaining water.

**Setup:** Water lost = rate × time = 0.5 gal/day × 30 days = 15 gallons.

**Computation:** Remaining water = 200 − 15 = **185 gallons**.

**Checking other choices:** Choosing 198.5 (D) would be the error of subtracting only 0.5 gallons (one day's evaporation), and 199.85 (E) makes no sense here. 140 (A) would come from subtracting 2 gallons/day, and 170 (B) from subtracting 1 gallon/day.

ANSWER 1: C

# Problem 2:
**Restating:** We're given a new operation where placing four numbers a, b, c, d in a box means compute a·d − b·c. Evaluate 3, 1, 4, 2 in the box.

**Setup:** Match a = 3, b = 1, c = 4, d = 2. The formula gives a·d − b·c.

**Computation:** 3·2 − 1·4 = 6 − 4 = **2**.

**Checking other choices:** A common error is computing a·b − c·d = 3 − 8 = −5 or a·c − b·d = 12 − 2 = 10; neither appears. Computing a·c − b·d reversed, i.e., b·d − a·c = 2 − 12 = −10, also doesn't appear. Carefully following the given formula: outer numbers multiply, inner numbers multiply, then subtract.

ANSWER 2: E

# Problem 3:
**Restating:** Find the sum of the prime factors of 2010.

**Setup (prime factorization):**
- 2010 is even: 2010 = 2 × 1005
- 1005 ends in 5: 1005 = 5 × 201
- 201 = 3 × 67
- 67 is prime.

So 2010 = 2 × 3 × 5 × 67.

**Computation:** Sum = 2 + 3 + 5 + 67 = **77**.

**Checking other choices:** 67 (A) is just the largest prime factor, not the sum. 75 (B) might come from adding 2 + 3 + 5 + 65 incorrectly. 201 (D) and 210 (E) are far too large — the sum of four small-ish primes can't approach 200.

ANSWER 3: C

# Problem 4:
**Restating:** Two 600 mL pitchers: one is 1/3 full of orange juice, the other 2/5 full. Both are topped off with water and combined. What fraction of the final mixture is orange juice?

**Setup:** 
- Orange juice in pitcher 1: (1/3) × 600 = 200 mL
- Orange juice in pitcher 2: (2/5) × 600 = 240 mL
- Total orange juice: 200 + 240 = 440 mL
- Total mixture: 600 + 600 = 1200 mL (both pitchers full when poured)

**Computation:** Fraction of OJ = 440/1200 = 44/120 = **11/30**.

**Checking other choices:** A common mistake is to average the fractions: (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — interestingly that gives the same answer here since both pitchers are the same size! Choosing 11/15 (E) forgets to account for the added water; 3/16 (B) or 1/8 (A) come from computation errors.

ANSWER 4: C

# Problem 5:
**Restating:** Each person starts with 10 and performs three operations in a specific order. Find who ends largest.

**Setup and computation:**
- **Jose:** subtract 1 → 9; double → 18; add 2 → **20**
- **Thuy:** double → 20; subtract 1 → 19; add 2 → **21**
- **Kareem:** subtract 1 → 9; add 2 → 11; double → **22**

**Comparing:** Kareem (22) > Thuy (21) > Jose (20).

**Checking other choices:** D and E are tempting because Thuy and Jose seem "close," but Kareem actually wins because the doubling happens *last* for him, magnifying the larger intermediate value 11. Order of operations matters!

ANSWER 5: C

# Problem 6:
**Restating:** Pizza has 12 slices. Peter eats 1 full slice plus half of another slice (shared equally with Paul). What fraction of the whole pizza did Peter eat?

**Setup:** 
- One full slice = 1/12 of the pizza
- Half of a slice = (1/2)(1/12) = 1/24 of the pizza

**Computation:** 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = **1/8**.

**Checking other choices:** 1/24 (A) counts only the half-slice; 1/12 (B) counts only the full slice; 1/6 (D) would be eating two full slices; 1/4 (E) is three slices.

ANSWER 6: C

# Problem 7:
**Restating:** Isabella takes four 100-point tests and wants an average of 95. Her first two scores are 97 and 91. What's the lowest possible third score that still allows her to reach her goal?

**Setup:** 
- Total points needed: 4 × 95 = 380
- Points so far: 97 + 91 = 188
- Points needed on last two tests: 380 − 188 = 192

**Key insight:** The maximum possible on the fourth test is 100. So the third test must be at least 192 − 100 = **92**.

**Verification:** If she scores 92 on test 3, she needs 100 on test 4: 97 + 91 + 92 + 100 = 380 → average 95. ✓ (Anything lower than 92 makes it impossible even with a perfect fourth test.)

**Checking other choices:** 90 (A) would require 102 on test 4 — impossible since the max is 100.

ANSWER 7: B

# Problem 8:
**Restating:** Daylight lasts 10 hours 24 minutes, sunrise is at 6:57 AM. The sunset time printed was wrong. Find the true sunset time.

**Setup:** Sunset = sunrise + length of daylight = 6:57 AM + 10 hours 24 minutes.

**Computation:** 
- 6:57 AM + 10 hours = 4:57 PM
- 4:57 PM + 24 minutes = **5:21 PM**

**Checking other choices:** 5:10 PM (A) would fit 10 hr 13 min; 5:41 PM (C) would fit 10 hr 44 min; 5:57 PM (D) forgets the 24 extra minutes (10 hours exactly). The wrong printed time (8:15 PM) was likely a typo — the true time must satisfy the correct daylight duration and sunrise.

ANSWER 8: B

# Problem 9:
**Restating:** The bike costs $500. Chandler receives $50 + $35 + $15 in birthday money and earns $16/week. How many weeks of paper route money (plus all birthday money) are needed to afford the bike?

**Setup:** 
- Birthday money: 50 + 35 + 15 = $100
- Still needed: 500 − 100 = $400
- Weeks needed: 400 ÷ 16 = **25 weeks**

**Checking other choices:** 24 (A) would leave him $16 short (only $484 total). Some might round up to 26 (C) unnecessarily, but 400 divides evenly by 16, so exactly 25 weeks suffices. Note: 16 × 25 = 400 exactly, so he has exactly $500 at 25 weeks.

ANSWER 9: B

# Problem 10:
**Restating:** 39 students each have a dog, a cat, or both. 20 have a dog, 26 have a cat. How many have both?

**Setup (inclusion-exclusion):** 
|Dog or Cat| = |Dog| + |Cat| − |Both|

**Computation:** 
39 = 20 + 26 − Both
39 = 46 − Both
Both = 46 − 39 = **7**

**Checking other choices:** 13 (B) would be (20+26)/2 − ... a common guess; 19 (C) might come from 39 − 20; 46

Problem 1:
The pool starts with 200 gallons. Water evaporates at 0.5 gallons per day, so in 30 days the amount evaporated is  
\[
0.5 \times 30 = 15 \text{ gallons}.
\]
Subtracting this from the original amount gives  
\[
200 - 15 = 185.
\]
So the pool has 185 gallons after 30 days. Choice D would result from evaporation of only 1.5 gallons, and Choice E would be for a very short time, so both are too high.

ANSWER 1: C

Problem 2:
The operation is defined by
\[
\begin{vmatrix}
a & c \\
b & d
\end{vmatrix}
= a \cdot d - b \cdot c.
\]
For
\[
\begin{vmatrix}
3 & 1 \\
4 & 2
\end{vmatrix},
\]
we have \(a=3\), \(c=1\), \(b=4\), and \(d=2\). Thus
\[
3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2.
\]
A common mistake is reversing the subtraction, but the formula specifically says \(ad-bc\).

ANSWER 2: E

Problem 3:
We need the sum of the prime factors of 2010. Factor:
\[
2010 = 2 \times 1005 = 2 \times 3 \times 335 = 2 \times 3 \times 5 \times 67.
\]
So the prime factors are \(2,3,5,67\). Their sum is
\[
2+3+5+67 = 77.
\]
Choice D, 201, is just a composite factor of 2010, not the sum of its prime factors.

ANSWER 3: C

Problem 4:
Each pitcher holds 600 mL.

First pitcher:
\[
\frac13 \times 600 = 200 \text{ mL of orange juice}.
\]
Second pitcher:
\[
\frac25 \times 600 = 240 \text{ mL of orange juice}.
\]
Total orange juice:
\[
200 + 240 = 440 \text{ mL}.
\]
The total mixture is the two full pitchers:
\[
600+600=1200 \text{ mL}.
\]
The fraction of the mixture that is orange juice is
\[
\frac{440}{1200}=\frac{11}{30}.
\]
This matches Choice C.

ANSWER 4: C

Problem 5:
Start with 10 for each person.

Jose:
\[
10-1=9,\quad 9\cdot 2=18,\quad 18+2=20.
\]

Thuy:
\[
10\cdot 2=20,\quad 20-1=19,\quad 19+2=21.
\]

Kareem:
\[
10-1=9,\quad 9+2=11,\quad 11\cdot 2=22.
\]

Kareem has the largest final answer, 22. The key is that Kareem doubles after adding 2, while Jose doubles before adding 2.

ANSWER 5: C

Problem 6:
The pizza has 12 slices. Peter ate 1 whole slice and half of another slice, so he ate
\[
1+\frac12=\frac32 \text{ slices}.
\]
The fraction of the whole pizza he ate is
\[
\frac{\frac32}{12}=\frac{3}{24}=\frac18.
\]
If someone counted only the full slice, they might get \(\frac1{12}\), but Peter also ate half of the shared slice.

ANSWER 6: C

Problem 7:
To average 95 on four 100-point tests, Isabella needs a total of
\[
4 \times 95 = 380.
\]
Her first two scores total
\[
97+91=188.
\]
Let her third score be \(x\). The fourth test can be at most 100, so to make the third score as low as possible, assume the fourth test is 100:
\[
188+x+100=380.
\]
Thus
\[
288+x=380,
\]
\[
x=92.
\]
If she scored 90 on the third test, she would need 102 on the fourth, which is impossible.

ANSWER 7: B

Problem 8:
The length of daylight is 10 hours 24 minutes. Sunrise was 6:57 AM, so sunset was that much later:
\[
6:57\text{ AM}+10\text{ hours}=4:57\text{ PM}.
\]
Then add 24 minutes:
\[
4:57\text{ PM}+24\text{ minutes}=5:21\text{ PM}.
\]
So the sun really set at 5:21 PM. The listed sunset of 8:15 PM is too late.

ANSWER 8: B

Problem 9:
Chandler needs \$500. His birthday money is
\[
50+35+15=100.
\]
He still needs
\[
500-100=400.
\]
He earns \$16 per week, so the number of weeks required is
\[
400\div 16=25.
\]
After 24 weeks he would have only \(16\times24=384\), plus \$100 for \$484, which is not enough.

ANSWER 9: B

Problem 10:
Let \(D\) be the number of students with a dog, \(C\) the number with a cat, and \(B\) the number with both. Every student has at least one of them, so by inclusion-exclusion:
\[
D+C-B=39.
\]
We know \(D=20\) and \(C=26\), so
\[
20+26-B=39.
\]
Thus
\[
46-B=39,
\]
\[
B=7.
\]
Choice E, 46, is just the sum of the dog and cat counts, but it overcounts the students with both.

ANSWER 10: A

Problem 1:
Water lost in 30 days: 0.5 × 30 = 15 gallons.
Remaining: 200 − 15 = 185 gallons.
(Tempting wrong answers come from subtracting only 0.5, or 1.5, or 30 gallons.)

ANSWER 1: C

Problem 2:
The notation is the 2×2 array with first row a, b and second row c, d, defined as a·d − b·c.
Here a = 3, b = 1, c = 4, d = 2.
Value = 3·2 − 1·4 = 6 − 4 = 2.
(Getting −2 means the terms were subtracted in the wrong order.)

ANSWER 2: E

Problem 3:
Factor 2010: 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67.
All of 2, 3, 5, 67 are prime.
Sum = 2 + 3 + 5 + 67 = 77.
(67 is just the largest factor; 201 and 210 are not sums of primes here.)

ANSWER 3: C

Problem 4:
Pitcher 1 orange juice: (1/3)(600) = 200 mL.
Pitcher 2 orange juice: (2/5)(600) = 240 mL.
Total orange juice = 200 + 240 = 440 mL.
Total liquid (both pitchers filled completely) = 600 + 600 = 1200 mL.
Fraction = 440/1200 = 11/30.
(11/15 would be the answer if the total were only one pitcher; averaging 1/3 and 2/5 incorrectly gives other choices.)

ANSWER 4: C

Problem 5:
Jose: (10 − 1) = 9, ×2 = 18, +2 = 20.
Thuy: 10 × 2 = 20, −1 = 19, +2 = 21.
Kareem: (10 − 1) = 9, +2 = 11, ×2 = 22.
Largest is Kareem's 22.

ANSWER 5: C

Problem 6:
Peter ate 1 full slice plus half of another slice = 1.5 slices.
Fraction of pizza = 1.5/12 = 3/24 = 1/8.
(1/12 counts only the whole slice; 1/24 counts only the half slice.)

ANSWER 6: C

Problem 7:
For an average of 95 over four tests, total needed = 4 × 95 = 380.
First two tests: 97 + 91 = 188, so tests 3 and 4 must total 380 − 188 = 192.
Best possible fourth score is 100, so third score ≥ 192 − 100 = 92.
Lowest possible third score = 92.
(90 would leave a needed 102 on the last test — impossible.)

ANSWER 7: B

Problem 8:
Sunrise 6:57 AM plus 10 hours = 4:57 PM.
Add 24 minutes: 4:57 PM + 24 min = 5:21 PM.
So the sun really set at 5:21 PM.

ANSWER 8: B

Problem 9:
Birthday money: 50 + 35 + 15 = 100 dollars.
Still needed: 500 − 100 = 400 dollars.
Weeks at 16 dollars/week: 400 ÷ 16 = 25 weeks exactly.
(25 weeks gives exactly 500 dollars, so no extra week is needed.)

ANSWER 9: B

Problem 10:
Let x = number with both. By inclusion–exclusion: 20 + 26 − x = 39.
46 − x = 39, so x = 7.

ANSWER 10: A

Problem 1:

We are asked how much water remains after 30 days.

- Starting amount: \(200\) gallons
- Evaporation rate: \(0.5\) gallons per day
- Number of days: \(30\)

First find the total amount evaporated:

\[
0.5 \times 30 = 15
\]

Now subtract this from the original 200 gallons:

\[
200-15=185
\]

So there will be \(185\) gallons left.

Choice A, \(140\), subtracts far too much.  
Choice B, \(170\), would mean 30 gallons evaporated.  
Choices D and E subtract much too little.

ANSWER 1: C

Problem 2:

The operation is defined by

\[
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=a\cdot d-b\cdot c.
\]

For the given numbers, \(a=3\), \(b=1\), \(c=4\), and \(d=2\).

Substitute into the rule:

\[
3\cdot 2-1\cdot 4
\]

\[
=6-4
\]

\[
=2
\]

Thus the value is \(2\).

A common mistake is to add the products, getting \(10\), but the rule says to subtract.

ANSWER 2: E

Problem 3:

We need the sum of the prime factors of \(2010\).

Factor \(2010\):

\[
2010=201\cdot 10
\]

\[
201=3\cdot 67,\qquad 10=2\cdot 5
\]

So

\[
2010=2\cdot 3\cdot 5\cdot 67.
\]

Its prime factors are \(2,3,5,\) and \(67\). Their sum is

\[
2+3+5+67=77.
\]

Thus the answer is \(77\).

Choice A, \(67\), includes only the largest prime factor.  
Choice E, \(210\), is not the sum of the prime factors.

ANSWER 3: C

Problem 4:

Each pitcher holds \(600\) mL.

The first pitcher is \(\frac13\) full of orange juice:

\[
\frac13\cdot 600=200\text{ mL}.
\]

The second pitcher is \(\frac25\) full of orange juice:

\[
\frac25\cdot 600=240\text{ mL}.
\]

So the total orange juice is

\[
200+240=440\text{ mL}.
\]

After both pitchers are filled with water, there are two full pitchers:

\[
600+600=1200\text{ mL}
\]

of mixture altogether.

Therefore, the fraction that is orange juice is

\[
\frac{440}{1200}.
\]

Simplify by dividing numerator and denominator by \(40\):

\[
\frac{440}{1200}=\frac{11}{30}.
\]

Choice D, \(\frac{11}{19}\), incorrectly compares orange juice with water instead of with the entire mixture.

ANSWER 4: C

Problem 5:

We calculate each person’s result carefully, following the operations in order.

For Jose:

\[
10-1=9
\]

\[
9\cdot 2=18
\]

\[
18+2=20.
\]

Jose gets \(20\).

For Thuy:

\[
10\cdot 2=20
\]

\[
20-1=19
\]

\[
19+2=21.
\]

Thuy gets \(21\).

For Kareem:

\[
10-1=9
\]

\[
9+2=11
\]

\[
11\cdot 2=22.
\]

Kareem gets \(22\), which is the largest result.

The operations are not interchangeable: doubling after adding 2, as Kareem does, makes that extra 2 become 4 after doubling.

ANSWER 5: C

Problem 6:

Peter ate one whole slice. He also shared another slice equally with Paul, so Peter ate half of that slice.

Thus Peter ate

\[
1+\frac12=\frac32
\]

slices.

The pizza had 12 slices, so the fraction Peter ate is

\[
\frac{\frac32}{12}=\frac32\cdot\frac1{12}=\frac3{24}=\frac18.
\]

Choice B, \(\frac{1}{12}\), counts only the one full slice and ignores Peter’s half of the shared slice.

ANSWER 6: C

Problem 7:

Isabella wants an average of \(95\) on four tests.

Therefore, she needs a total of

\[
4\cdot 95=380
\]

points.

Her first two scores total

\[
97+91=188.
\]

So her third and fourth scores together must total

\[
380-188=192.
\]

To make her third-test score as low as possible, she would need to earn the maximum possible score, \(100\), on the fourth test.

Thus her lowest possible third-test score is

\[
192-100=92.
\]

If she scored only \(90\) on the third test, even a \(100\) on the fourth would give only \(190\) more points, not the needed \(192\).

ANSWER 7: B

Problem 8:

The sunrise was at \(6{:}57\) AM, and daylight lasted \(10\) hours and \(24\) minutes.

Add 10 hours:

\[
6{:}57\text{ AM}+10\text{ hours}=4{:}57\text{ PM}.
\]

Now add 24 minutes:

\[
4{:}57\text{ PM}+24\text{ minutes}=5{:}21\text{ PM}.
\]

So the sun really set at \(5{:}21\) PM.

The listed sunset of \(8{:}15\) PM would give more than 13 hours of daylight, so it cannot be correct.

ANSWER 8: B

Problem 9:

Chandler receives birthday money from three relatives:

\[
50+35+15=100
\]

dollars.

The bike costs \(500\) dollars, so he still needs

\[
500-100=400
\]

dollars.

He earns \(16\) dollars per week. The number of weeks needed is

\[
400\div 16=25.
\]

After 25 weeks, he earns

\[
25\cdot 16=400
\]

dollars, exactly enough to buy the bike.

Choice A, 24 weeks, would earn only

\[
24\cdot 16=384,
\]

which is not enough.

ANSWER 9: B

Problem 10:

There are 39 students total, and every student has a dog, a cat, or both.

- Students with dogs: \(20\)
- Students with cats: \(26\)

If we add these, we get

\[
20+26=46.
\]

However, students who have both a dog and a cat were counted twice. Since there are only 39 students altogether, the number counted twice is

\[
46-39=7.
\]

Therefore, \(7\) students have both a dog and a cat.

Choice E, \(46\), is the total after double-counting the students who have both pets.

ANSWER 10: A