🔬 Model Lab

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C

Problem 1:
We are asked for Joe's total time home to school. He walks half the distance, then runs half the distance.

Let the one-way distance be $D$. Let walking speed be $v$.

Walk half-way: distance $D/2$ at speed $v$ takes 6 minutes given.
So $(D/2)/v = 6$.

Run half-way: distance $D/2$ at speed $3v$ because he runs 3 times as fast as he walks.
Time to run = distance / speed = $(D/2)/(3v)$.

This is $1/3$ of the walking time for the same distance:
$(D/2)/(3v) = (1/3)\times (D/2)/v = (1/3)\times 6 = 2$ minutes.

Total time = walk time + run time = $6+2 = 8$ minutes.

Wrong choices like 7 would be $6+1$, assuming run is even faster, and 7.3, 7.7, 8.3 come from mis-dividing. The exact calculation gives 8.
ANSWER 1: D

Problem 2:
We need number made in shots 21 to 25.

After 20 shots, made $55\%$:
$0.55 \times 20 = 11$ shots made. This must be an integer, and it is.

After 25 shots total, made $56\%$:
$0.56 \times 25 = 14$ shots made. Since $56/100 \times 25 = 14$, integer.

So increase in makes = $14 - 11 = 3$.
That means of the last 5 shots, she made 3.

Check: $11+3=14$ out of $25 = 56\%$. Works. Choices 1,2,4,5 would not give 56%.
ANSWER 2: C

Problem 3:
We need average inches per hour = total inches divided by total hours.

July has 31 days. Each day has 24 hours.
So total hours in July = $31 \times 24$.

Total rain = 366 inches.
Average = $366 \div (31 \times 24) = 366/(31 \times 24)$.

Choices B, C, E multiply where they should divide, giving huge numbers. Choice D is hours per inch, upside down. Only A is total divided by total time.
ANSWER 3: A

Problem 4:
We match 3 adult photos to 3 baby photos at random. What is probability all 3 matches are correct?

Think of ordering baby pictures. For first celebrity, 3 choices of baby picture, only 1 correct. But events are not independent.

Count equally likely matchings: number of permutations of 3 items is $3! = 3\times 2\times 1 = 6$.
Only 1 of those 6 is fully correct.

So probability = $1/6$.

$1/9$ would be tempting if you thought $1/3 \times 1/3$, but after matching one, choices reduce. $1/3$ is probability of getting a particular one right, not all three.
ANSWER 4: B

Problem 5:
We compare two orders of operations: add 6% tax and subtract 20% discount.

Let price $P=90$.
Jack: $(P \times 1.06)\times 0.80$.
Compute: $90\times 1.06 = 95.40$. Then $95.40\times 0.80 = 76.32$.

Jill: $(P \times 0.80)\times 1.06$.
Compute: $90\times 0.80 = 72.00$. Then $72.00\times 1.06 = 76.32$.

They are equal because multiplication commutes:
$P\times 1.06 \times 0.8 = P \times 0.8 \times 1.06$.

So Jack's total minus Jill's total = $76.32-76.32 = \$0$.

Non-zero choices come from arithmetic error or thinking order matters for percent.
ANSWER 5: C

Problem 6:
Given February 13 is Friday, find day of week of February 1.

Difference: $13-1 = 12$ days later.
$12 \pmod 7 = 5$, so Feb 1 is 5 weekdays before Friday.

Count back: Friday minus 1 = Thursday (Feb 12), minus 2 = Wednesday (Feb 11), minus 3 = Tuesday (Feb 10), minus 4 = Monday (Feb 9), minus 5 = Sunday (Feb 8)? Wait continue to Feb 1.

Better: Feb 13 Friday => Feb 6 is also Friday, since $13-7=6$.
Then: Feb 6 Friday, Feb 5 Thursday, Feb 4 Wednesday, Feb 3 Tuesday, Feb 2 Monday, Feb 1 Sunday.

So Feb 1 is Sunday.

Monday would be Feb 2. Saturday, etc. are off by miscounting.
ANSWER 6: A

Problem 7:
Pool capacity 24,000 gallons. 4 hoses at 2.5 gallons per minute each. How many hours?

Rate per hose = 2.5 gal/min.
Total rate = $4 \times 2.5 = 10$ gal/min.

Time in minutes = volume / rate = $24000 / 10 = 2400$ minutes.

Convert to hours: divide by 60: $2400/60 = 40$ hours.

Choices 42,44,46,48 would come from forgetting 4 hoses or mis-converting minutes to hours.
ANSWER 7: A

Problem 8:
Photo 8 by 10. Frame adds 2-inch border on all sides. Find border area.

Photo area = $8 \times 10 = 80$.

Outer rectangle: height = $8 + 2 + 2 = 12$ (2 inches top and bottom). Width = $10 + 2 + 2 = 14$ (2 inches left and right).
Outer area = $12 \times 14 = 168$.

Border = outer minus inner = $168 - 80 = 88$ square inches.

36,40 would be perimeter-like mistakes; 72,64 mis-add border only once. Correct inclusion of both sides gives 88.
ANSWER 8: E

Problem 9:
Trip 100 miles, 3 hours total including lunch. Driving speed 40 mph. Find lunch length in minutes.

Driving time = distance / speed = $100/40 = 2.5$ hours.
$2.5$ hours = $2.5 \times 60 = 150$ minutes.

Total elapsed = 3 hours = 180 minutes.
Lunch = total - driving = $180 - 150 = 30$ minutes.

15,45,60 would come from wrong driving time. 30 fits exactly.
ANSWER 9: B

Problem 10:
Each starts with 10 but applies operations in different order. Compute each final.

Jose: subtract 1, double, add 2:
$(10-1)=9$; $9\times 2=18$; $18+2=20$.

Thuy: double, subtract 1, add 2:
$10\times 2=20$; $20-1=19$; $19+2=21$.

Kareem: subtract 1, add 2, double:
$(10-1)=9$; $9+2=11$; $11\times 2=22$.

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

So not Jose, not Thuy, not ties.
ANSWER 10: C

Problem 1:
The problem asks for the total time it took Joe to travel from home to school.
The distance from home to school is divided into two equal halves:
- The first half was walked in 6 minutes.
- The second half was run. Since Joe ran 3 times as fast as he walked, covering the same distance takes $\frac{1}{3}$ of the time it took to walk.
Therefore, the time spent running is:
$$\frac{6\text{ minutes}}{3} = 2\text{ minutes}$$
The total time for the trip is the sum of the walking time and running time:
$$6 + 2 = 8\text{ minutes}$$

ANSWER 1: D

---

Problem 2:
The problem asks how many of the last 5 shots Sally made.
- First, after 20 shots, Sally made $55\%$ of them:
$$20 \times 0.55 = 11\text{ shots made}$$
- After taking 5 more shots, she has taken a total of $20 + 5 = 25$ shots.
- Her new percentage is $56\%$, so the total number of shots made is:
$$25 \times 0.56 = 25 \times \frac{56}{100} = \frac{56}{4} = 14\text{ shots made}$$
- The number of shots she made out of the last 5 is the difference in total shots made:
$$14 - 11 = 3\text{ shots}$$

ANSWER 2: C

---

Problem 3:
The problem asks for the average rainfall in inches per hour in Cherrapunji, India, during July 1861.
- Total rainfall = $366$ inches.
- The month of July has $31$ days.
- Each day has $24$ hours.
- Thus, the total number of hours in July is $31 \times 24$.
To find the average rainfall in inches per hour, divide the total rainfall by the total number of hours:
$$\text{Average rainfall} = \frac{366}{31 \times 24}$$
This matches Choice A.

ANSWER 3: A

---

Problem 4:
The problem asks for the probability that a reader guessing at random will match all three celebrities with their baby pictures correctly.
- There are 3 celebrities and 3 baby pictures.
- The number of possible one-to-one matchings (permutations) of the 3 baby pictures to the 3 celebrities is:
$$3! = 3 \times 2 \times 1 = 6$$
- Exactly 1 of these permutations represents the completely correct matching.
- Since the reader is guessing at random, each of the 6 matchings is equally likely.
Thus, the probability of getting all three correct is:
$$\frac{1}{6}$$

ANSWER 4: B

---

Problem 5:
The problem asks for the difference between Jack's calculated total and Jill's calculated total.
Let the initial price be $P = \$90.00$.
- Jack's calculation:
  1. Adds $6\%$ sales tax: $P \times (1 + 0.06) = P \times 1.06$
  2. Subtracts $20\%$ discount: $(P \times 1.06) \times (1 - 0.20) = P \times 1.06 \times 0.80$
- Jill's calculation:
  1. Subtracts $20\%$ discount: $P \times (1 - 0.20) = P \times 0.80$
  2. Adds $6\%$ sales tax: $(P \times 0.80) \times (1 + 0.06) = P \times 0.80 \times 1.06$

By the commutative property of multiplication:
$$P \times 1.06 \times 0.80 = P \times 0.80 \times 1.06$$
Jack's total and Jill's total are mathematically identical:
$$\text{Jack's total} - \text{Jill's total} = \$0$$

ANSWER 5: C

---

Problem 6:
The problem asks for the day of the week of February 1, given that February 13 is a Friday.
The number of days between February 1 and February 13 is:
$$13 - 1 = 12\text{ days}$$
We determine the day of the week by looking at $12 \pmod 7$:
$$12 = 1 \times 7 + 5$$
Counting back 12 days (or counting back 5 days, which is equivalent to counting forward 2 days):
- 1 day before Friday is Thursday
- 2 days before is Wednesday
- 3 days before is Tuesday
- 4 days before is Monday
- 5 days before is Sunday

Alternatively, subtracting 7 days from Friday the 13th gives Friday the 6th, and counting backwards:
- Feb 6: Friday
- Feb 5: Thursday
- Feb 4: Wednesday
- Feb 3: Tuesday
- Feb 2: Monday
- Feb 1: Sunday

Therefore, February 1 is a Sunday.

ANSWER 6: A

---

Problem 7:
The problem asks for the time, in hours, required to fill a 24,000-gallon pool using 4 hoses.
- Each hose supplies $2.5$ gallons per minute.
- Together, the 4 hoses supply:
$$4 \times 2.5 = 10\text{ gallons per minute}$$
- In one hour (60 minutes), the 4 hoses supply:
$$10 \times 60 = 600\text{ gallons per hour}$$
- To fill the 24,000-gallon pool, the number of hours needed is:
$$\frac{24,000}{600} = \frac{240}{6} = 40\text{ hours}$$

ANSWER 7: A

---

Problem 8:
The problem asks for the area of a 2-inch border surrounding a rectangular photograph.
- The photograph has dimensions:
$$\text{Height} = 8\text{ inches}, \quad \text{Width} = 10\text{ inches}$$
- The area of the photograph alone is:
$$8 \times 10 = 80\text{ square inches}$$
- The border adds 2 inches to every side (top, bottom, left, and right).
- The outer dimensions of the frame are:
$$\text{Outer height} = 8 + 2 + 2 = 12\text{ inches}$$
$$\text{Outer width} = 10 + 2 + 2 = 14\text{ inches}$$
- The total area of the frame including the photograph is:
$$12 \times 14 = 168\text{ square inches}$$
- The area of the border is the total area minus the area of the photograph:
$$\text{Border area} = 168 - 80 = 88\text{ square inches}$$

ANSWER 8: E

---

Problem 9:
The problem asks for the duration of Casey's lunch break, in minutes.
- The total distance of the trip was $100$ miles.
- Her driving speed was $40$ miles per hour.
- The time spent driving is:
$$\text{Driving time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{40} = 2.5\text{ hours}$$
- The total trip took $3$ hours.
- Therefore, the time spent on the lunch break was:
$$\text{Break time} = 3 - 2.5 = 0.5\text{ hours}$$
- Converting $0.5$ hours to minutes:
$$0.5 \times 60 = 30\text{ minutes}$$

ANSWER 9: B

---

Problem 10:
The problem asks who gets the largest final answer among Jose, Thuy, and Kareem.
Each person starts with the number 10:

1. **Jose**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Doubles the answer: $9 \times 2 = 18$
   - Adds 2: $18 + 2 = 20$

2. **Thuy**:
   - Doubles 10: $10 \times 2 = 20$
   - Subtracts 1: $20 - 1 = 19$
   - Adds 2: $19 + 2 = 21$

3. **Kareem**:
   - Subtracts 1 from 10: $10 - 1 = 9$
   - Adds 2 to the number: $9 + 2 = 11$
   - Doubles the result: $11 \times 2 = 22$

Comparing the results:
- Jose: 20
- Thuy: 21
- Kare

# Problem 1:
**What's asked:** Total time from home to school, given walking half takes 6 min and running is 3× walking speed.

**Setup:** The two halves are the same distance. Speed triples, so time for the same distance is divided by 3.

**Computation:**
- Walking second half would take 6 minutes; running takes 6 ÷ 3 = **2 minutes**.
- Total time = 6 + 2 = **8 minutes**.

**Rule out wrong choices:** 7 or 7.3 would mean the run was faster than 3×; 8.3 or 7.7 would mean the run took almost as long as the walk.

ANSWER 1: D

# Problem 2:
**What's asked:** Number of successful shots among the last 5.

**Setup:** Compute made shots before and after the extra 5 shots.

**Computation:**
- After 20 shots at 55%: 0.55 × 20 = **11 made**.
- After 25 shots at 56%: 0.56 × 25 = **14 made**.
- Last 5 shots made: 14 − 11 = **3**.

**Rule out wrong choices:** 4 would give 15/25 = 60%, not 56%; 2 would give 13/25 = 52%.

ANSWER 2: C

# Problem 3:
**What's asked:** Average rainfall per hour, in inches.

**Setup:** July has 31 days × 24 hours/day = 744 total hours. Average = total rainfall ÷ total hours.

**Computation:** Average = 366 ÷ (31 × 24).

**Rule out wrong choices:** (B), (C), (E) multiply by days/hours instead of dividing; (D) inverts the fraction (that would be hours per inch).

ANSWER 3: A

# Problem 4:
**What's asked:** Probability of matching all 3 celebrities to their baby photos by random guessing.

**Setup:** A random guess is a random arrangement (permutation) of the 3 baby photos among the 3 celebrities.

**Computation:**
- Number of possible arrangements: 3! = 3 × 2 × 1 = **6**.
- Exactly 1 arrangement is fully correct.
- Probability = 1/6.

**Rule out wrong choices:** 1/3 is the chance of matching just one celebrity correctly; 1/9 assumes independent 1/3 chances, but the matches are not independent.

ANSWER 4: B

# Problem 5:
**What's asked:** Jack's total minus Jill's total.

**Setup:**
- Jack: $90 → add 6% tax → subtract 20%.
- Jill: $90 → subtract 20% → add 6% tax.

**Computation:**
- Jack: 90 × 1.06 = 95.40; then 95.40 × 0.80 = **$76.32**.
- Jill: 90 × 0.80 = 72.00; then 72 × 1.06 = **$76.32**.
- Jack − Jill = 76.32 − 76.32 = **$0**.

**Rule out wrong choices:** ±$0.53 and ±$1.06 trap you into thinking order matters — but multiplying by 1.06 and 0.80 in either order gives the same product (commutativity).

ANSWER 5: C

# Problem 6:
**What's asked:** Day of week of February 1, given Friday the 13th falls in that February.

**Setup:** Feb 1 is 12 days before Feb 13. Since 12 = 7 + 5, Feb 1 is **5 days of the week** earlier than Friday.

**Computation:** Counting back 5 days from Friday: Thu (1), Wed (2), Tue (3), Mon (4), **Sun (5)**.

**Rule out wrong choices:** Thursday would make the 13th a Saturday; Monday would make it a Thursday; etc. Check: Feb 1 Sunday → Feb 8 Sunday → Feb 13 Friday. ✓

ANSWER 6: A

# Problem 7:
**What's asked:** Hours to fill the pool.

**Setup:** 4 hoses × 2.5 gallons/minute each gives the total fill rate.

**Computation:**
- Combined rate: 4 × 2.5 = 10 gallons/minute.
- Time: 24,000 ÷ 10 = 2,400 minutes.
- Convert: 2,400 ÷ 60 = **40 hours**.

**Rule out wrong choices:** 48 hours would result from using only one hose's rate of 2.5 gal/min incorrectly; forgetting the hours conversion would give 2,400.

ANSWER 7: A

# Problem 8:
**What's asked:** Area of the border (frame) around the photo.

**Setup:** The border adds 2 inches on *each* side, so each dimension grows by 2 + 2 = 4 inches.

**Computation:**
- Frame outer dimensions: (8 + 4) × (10 + 4) = 12 × 14 → area = 168 sq in.
- Photo area: 8 × 10 = 80 sq in.
- Border area: 168 − 80 = **88 sq in**.

**Rule out wrong choices:** 64 comes from adding only 2 total to each dimension (10 × 12 = 144 − 80); 40 comes from 2 × perimeter-style errors.

ANSWER 8: E

# Problem 9:
**What's asked:** Length of the lunch break in minutes.

**Setup:** Total time = driving time + break time. Driving time = distance ÷ driving speed.

**Computation:**
- Driving time: 100 miles ÷ 40 mph = 2.5 hours.
- Break time: 3 − 2.5 = 0.5 hours = **30 minutes**.

**Rule out wrong choices:** 60 minutes would mean only 2 hours of driving (50 mph average); 15 would mean 2.75 hours driving (~36.4 mph).

ANSWER 9: B

# Problem 10:
**What's asked:** Who ends with the largest number after each performs their operations on 10.

**Setup:** Carefully track the order of operations for each person.

**Computation:**
- Jose: 10 − 1 = 9; 9 × 2 = 18; 18 + 2 = **20**.
- Thuy: 10 × 2 = 20; 20 − 1 = 19; 19 + 2 = **21**.
- Kareem: 10 − 1 = 9; 9 + 2 = 11; 11 × 2 = **22**.

Kareem's 22 is the largest. (Doubling last is best because it amplifies the +2.)

**Rule out wrong choices:** Jose is smallest (20); "Thuy and Kareem" fails since Thuy's 21 < Kareem's 22.

ANSWER 10: C

Problem 1:
We need the total time from home to school. Joe walked the first half in 6 minutes. The second half is the same distance, but he ran 3 times as fast. Since time is inversely proportional to speed, the time to run the same distance is

\[
\frac{6}{3}=2 \text{ minutes}.
\]

So the total time is

\[
6+2=8 \text{ minutes}.
\]

The tempting wrong answer would be to add some extra time incorrectly, but the running half takes exactly one-third as long as the walking half.

ANSWER 1: D

Problem 2:
Sally made 55% of her first 20 shots:

\[
0.55 \times 20 = 11.
\]

After 5 more shots, she has taken

\[
20+5=25
\]

shots total, and made 56% of them:

\[
0.56 \times 25 = 14.
\]

So during the last 5 shots, she made

\[
14-11=3
\]

shots.

ANSWER 2: C

Problem 3:
The average rainfall in inches per hour is total rainfall divided by total hours.

July has 31 days, and each day has 24 hours, so July has

\[
31 \times 24
\]

hours.

The average rainfall per hour is therefore

\[
\frac{366}{31 \times 24}.
\]

ANSWER 3: A

Problem 4:
There are 3 celebrities and 3 baby photos. A random reader is making a one-to-one matching. The total number of possible matchings is the number of permutations of 3 objects:

\[
3! = 3 \times 2 \times 1 = 6.
\]

Only one of those matchings is completely correct. Therefore the probability is

\[
\frac{1}{6}.
\]

A tempting wrong choice is \(\frac{1}{9}\), but that would treat the three matches as independent, while the reader cannot assign the same baby photo to two celebrities.

ANSWER 4: B

Problem 5:
Jack first adds 6% tax to $90, then subtracts 20%:

\[
90(1.06)(0.80).
\]

Jill first subtracts 20% from $90, then adds 6% tax:

\[
90(0.80)(1.06).
\]

Multiplication is commutative, so these are equal:

\[
90(1.06)(0.80)=90(0.80)(1.06).
\]

Thus Jack’s total minus Jill’s total is

\[
0.
\]

ANSWER 5: C

Problem 6:
If February 13 is a Friday, then February 1 is 12 days earlier. Since 12 days is 1 week plus 5 days, February 1 is 5 days before Friday.

Counting back 5 days from Friday:

- 1 day before Friday: Thursday
- 2 days before: Wednesday
- 3 days before: Tuesday
- 4 days before: Monday
- 5 days before: Sunday

So February 1 is a Sunday.

ANSWER 6: A

Problem 7:
Each hose supplies 2.5 gallons per minute, and there are 4 hoses, so the total rate is

\[
4 \times 2.5 = 10 \text{ gallons per minute}.
\]

The pool holds 24,000 gallons, so the time in minutes is

\[
\frac{24000}{10}=2400 \text{ minutes}.
\]

Convert to hours:

\[
\frac{2400}{60}=40 \text{ hours}.
\]

ANSWER 7: A

Problem 8:
The photograph has area

\[
8 \times 10 = 80 \text{ square inches}.
\]

The frame adds a 2-inch border on all sides, so the outer dimensions are

\[
8+2+2=12
\]

inches high and

\[
10+2+2=14
\]

inches wide.

The outer area is

\[
12 \times 14 = 168.
\]

The border area is the outer area minus the photograph area:

\[
168-80=88.
\]

ANSWER 8: E

Problem 9:
Casey drove 100 miles at an average driving speed of 40 miles per hour. The driving time is

\[
\frac{100}{40}=2.5 \text{ hours}.
\]

Convert to minutes:

\[
2.5 \times 60 = 150 \text{ minutes}.
\]

The entire trip took 3 hours, which is

\[
3 \times 60 = 180 \text{ minutes}.
\]

So the lunch break lasted

\[
180-150=30 \text{ minutes}.
\]

ANSWER 9: B

Problem 10:
Compute each person’s final number.

Jose:

\[
10-1=9,
\]
\[
2 \times 9=18,
\]
\[
18+2=20.
\]

Thuy:

\[
2 \times 10=20,
\]
\[
20-1=19,
\]
\[
19+2=21.
\]

Kareem:

\[
10-1=9,
\]
\[
9+2=11,
\]
\[
2 \times 11=22.
\]

Kareem’s final answer, 22, is the largest.

ANSWER 10: C

Problem 1:
Joe walks the first half in 6 minutes. Running is 3 times as fast, so covering the same distance (the second half) takes 6 ÷ 3 = 2 minutes.
Total time = 6 + 2 = 8 minutes.
(Choices like 7.3 or 7.7 would come from wrongly dividing the *total* trip time or averaging speeds; the correct approach divides only the second half's time by 3.)

ANSWER 1: D

Problem 2:
Shots made in first 20: 55% of 20 = 0.55 × 20 = 11.
Shots made in first 25: 56% of 25 = 0.56 × 25 = 14.
So in the last 5 shots she made 14 − 11 = 3.
(Tempting error: thinking a 1% rise means only 1 shot — but the denominator changed too.)

ANSWER 2: C

Problem 3:
"Average per hour" = total rain ÷ total hours. July has 31 days, each with 24 hours, so hours = 31 × 24.
Average = 366 / (31 × 24).
Multiplying instead of dividing (B, C, E) gives units of inches·hours, which is meaningless; D is the reciprocal (hours per inch).

ANSWER 3: A

Problem 4:
There are 3! = 6 equally likely ways to match 3 baby photos to 3 celebrities, and exactly 1 is completely correct.
Probability = 1/6.
(1/9 would come from wrongly treating the three guesses as independent 1/3 events; the matches are dependent.)

ANSWER 4: B

Problem 5:
Jack: 90 × 1.06 × 0.80 = 90 × 0.848 = $76.32.
Jill: 90 × 0.80 × 1.06 = 90 × 0.848 = $76.32.
Multiplication is commutative, so the order of applying the tax and the discount doesn't matter.
Difference = 76.32 − 76.32 = $0.

ANSWER 5: C

Problem 6:
Days that are the same weekday differ by multiples of 7. From Friday the 13th, go back 7 days: February 6 is a Friday. Then:
Feb 6 Fri, Feb 5 Thu, Feb 4 Wed, Feb 3 Tue, Feb 2 Mon, Feb 1 Sun.
(Check: 13 − 1 = 12, and 12 = 7 + 5, so Feb 1 is 5 weekdays before Friday → Sunday.)

ANSWER 6: A

Problem 7:
Combined rate = 4 × 2.5 = 10 gallons per minute = 10 × 60 = 600 gallons per hour.
Time = 24,000 ÷ 600 = 40 hours.
(Forgetting to combine all four hoses or mixing minutes and hours produces the other choices.)

ANSWER 7: A

Problem 8:
Photo: 8 by 10, area = 80 square inches.
With a 2-inch border on all sides, the outer dimensions are (8 + 2 + 2) by (10 + 2 + 2) = 12 by 14, area = 168.
Border area = 168 − 80 = 88 square inches.
(Adding only 2 inches total instead of 2 on each side gives 10 × 12 = 120 − 80 = 40, choice B — a common trap.)

ANSWER 8: E

Problem 9:
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours.
Lunch break = total time − driving time = 3 − 2.5 = 0.5 hour = 30 minutes.
(Note the 40 mph is her speed *while driving*, not overall, so we can't use 3 hours directly.)

ANSWER 9: B

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

ANSWER 10: C

Problem 1:

We are asked for Joe’s total travel time.

Joe walked the first half of the distance in 6 minutes. Let the distance from home to school be \(2d\), so each half is \(d\).

- Walking half: \(6\) minutes.
- He runs 3 times as fast as he walks, so he takes \(\frac13\) as much time to cover the same distance.
- Running the other half takes
  \[
  6\div 3=2\text{ minutes}.
  \]

Thus his total time is
\[
6+2=8\text{ minutes}.
\]

A tempting error is to divide the entire 6 minutes by 3, but only the second half is run at the faster speed.

ANSWER 1: D

Problem 2:

We are asked how many of Sally’s last 5 shots were made.

After 20 shots, Sally has made \(55\%\) of them:
\[
0.55(20)=11.
\]
So she has made 11 shots after the first 20.

After 5 more shots, she has taken
\[
20+5=25
\]
shots total. Her new percentage is \(56\%\), so she has made
\[
0.56(25)=14
\]
shots total.

Therefore, among the last 5 shots, she made
\[
14-11=3.
\]

A tempting wrong choice is 1, from thinking the percentage increased by only \(1\%\), but the number of total shots also increased.

ANSWER 2: C

Problem 3:

We are asked for the average rainfall per hour.

July has 31 days, and each day has 24 hours. Thus July contains
\[
31\times 24
\]
hours.

The total rainfall was 366 inches, so the average rainfall per hour is
\[
\frac{366}{31\times24}.
\]

The other expressions either multiply by the number of hours instead of dividing by it, or reverse the fraction.

ANSWER 3: A

Problem 4:

We are asked for the probability of correctly matching all three celebrities to their baby pictures by random guessing.

There are \(3!\) possible ways to match 3 celebrities with 3 baby pictures:
\[
3!=3\cdot2\cdot1=6.
\]

Only one of these 6 arrangements matches all three correctly. Therefore the probability is
\[
\frac16.
\]

Equivalently:

- The first celebrity has probability \(\frac13\) of being matched correctly.
- Then the second has probability \(\frac12\).
- The third is then forced to be correct.

So:
\[
\frac13\cdot\frac12=\frac16.
\]

A tempting answer is \(\frac19\), obtained by incorrectly treating all three choices as independent choices among three pictures.

ANSWER 4: B

Problem 5:

We are asked for Jack’s total minus Jill’s total.

The original coat price is \(\$90.00\).

Jack first adds \(6\%\) tax:
\[
90(1.06)=95.40.
\]
Then he takes off \(20\%\):
\[
95.40(0.80)=76.32.
\]

Jill first takes off \(20\%\):
\[
90(0.80)=72.00.
\]
Then she adds \(6\%\) tax:
\[
72(1.06)=76.32.
\]

Both clerks get the same total, so
\[
76.32-76.32=0.
\]

The percentages give the same result because multiplying by \(1.06\) and multiplying by \(0.80\) can be done in either order.

ANSWER 5: C

Problem 6:

We are asked what day of the week February 1 is if February 13 is Friday.

February 13 is 12 days after February 1:
\[
13-1=12.
\]

Since
\[
12=7+5,
\]
February 13 is 5 days after February 1.

Count backward 5 days from Friday:

- Thursday: 1 day before Friday
- Wednesday: 2 days before Friday
- Tuesday: 3 days before Friday
- Monday: 4 days before Friday
- Sunday: 5 days before Friday

Therefore, February 1 is Sunday.

A tempting mistake is to count 13 days rather than the 12-day gap from the 1st to the 13th.

ANSWER 6: A

Problem 7:

We are asked how many hours it takes to fill the pool.

Each of the 4 hoses supplies 2.5 gallons per minute, so together they supply
\[
4(2.5)=10
\]
gallons per minute.

The pool holds 24,000 gallons. The filling time in minutes is
\[
\frac{24000}{10}=2400\text{ minutes}.
\]

Convert minutes to hours:
\[
2400\div60=40.
\]

So it takes 40 hours.

A tempting error is to divide by 2.5 and forget that all four hoses are running at once.

ANSWER 7: A

Problem 8:

We are asked for the area of the border.

The photograph is 8 inches high and 10 inches wide. Since the border is 2 inches wide on every side, the outside dimensions of the frame are:

Height:
\[
8+2+2=12
\]

Width:
\[
10+2+2=14
\]

So the total framed area is
\[
12\cdot14=168.
\]

The photo itself has area
\[
8\cdot10=80.
\]

Therefore, the border area is
\[
168-80=88
\]
square inches.

A tempting wrong answer is 64, which is the area added by extending only one dimension incorrectly.

ANSWER 8: E

Problem 9:

We are asked for the length of Casey’s lunch break in minutes.

Casey drove 100 miles at an average driving speed of 40 miles per hour. Her driving time was
\[
\frac{100}{40}=2.5\text{ hours}.
\]

The entire trip, including lunch, took 3 hours. Thus lunch took
\[
3-2.5=0.5\text{ hours}.
\]

Convert to minutes:
\[
0.5(60)=30\text{ minutes}.
\]

A tempting answer is 40, but 40 is her speed in miles per hour, not the lunch time.

ANSWER 9: B

Problem 10:

We are asked who gets the largest final answer.

Jose:
\[
10-1=9
\]
\[
9\cdot2=18
\]
\[
18+2=20
\]

Thuy:
\[
10\cdot2=20
\]
\[
20-1=19
\]
\[
19+2=21
\]

Kareem:
\[
10-1=9
\]
\[
9+2=11
\]
\[
11\cdot2=22
\]

The results are:

- Jose: \(20\)
- Thuy: \(21\)
- Kareem: \(22\)

Kareem has the largest final answer. The order matters: adding 2 before doubling makes that added 2 become 4 after the doubling.

ANSWER 10: C