🔬 Model Lab

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D

Problem 1:
We are asked for time per mile now minus time per mile as a boy.

As a boy: 15 miles in 3 hours and 30 minutes.
3 hours 30 minutes = 3×60 + 30 = 180 + 30 = 210 minutes.
Time per mile = 210 ÷ 15 = 14 minutes per mile.

As an old man: 10 miles in 4 hours.
4 hours = 4×60 = 240 minutes.
Time per mile = 240 ÷ 10 = 24 minutes per mile.

Difference = 24 − 14 = 10 minutes longer now.

10 minutes is choice B. Choice A 6, C 15, D 18, E 30 do not match the calculation.
ANSWER 1: B

Problem 2:
We are asked to compute
1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All have common denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45.
45 is the sum 1 to 9.

So sum = (45 + 55)/10 = 100/10 = 10.

Choice A 4 1/2 would be just (1+...+9)/10, forgetting the 55/10. Choice C 9 and E 11 are near misses. The correct is 10, choice D.
ANSWER 2: D

Problem 3:
We are asked which number has the smallest prime factor. Find the smallest prime divisor of each:

A. 55 = 5 × 11, smallest prime factor 5.
B. 57 = 3 × 19, smallest prime factor 3.
C. 58 = 2 × 29, smallest prime factor 2.
D. 59 is prime, smallest prime factor 59 itself.
E. 61 is prime, smallest prime factor 61 itself.

Compare 5, 3, 2, 59, 61. The smallest of these is 2, from 58.

So choice C wins. Note 2 is the smallest possible prime, so no number can beat it.
ANSWER 3: C

Problem 4:
We need two primes p+q = 85, find p×q.

85 is odd. odd = even + odd. So one of the primes must be even and the other odd.

The only even prime is 2. So one number must be 2.

Then the other is 85 − 2 = 83. 83 is prime.

Product = 2 × 83 = 166.

This is choice E. The other choices: A 85, B 91, C 115, D 133 are products of other pairs summing to non-85 or non-prime.
ANSWER 4: E

Problem 5:
Lucius starts at 100 and subtracts 7 each time.

Term 1 = 100
Term 2 = 100 − 7 = 93
Term 3 = 100 − 2×7 = 86
In general term n = 100 − 7×(n−1).

For the 10th number, n=10:
10th = 100 − 7×9 = 100 − 63 = 37.

Choice B is 37. Choice A 30 would be the 11th term, 100−70.
ANSWER 5: B

Problem 6:
We have an 8 by 8 board of 64 unit squares. A square "does not touch the outer edge" means it is strictly interior.

Remove the outer border: 1 square thick all around. The interior is (8−2) by (8−2) = 6 by 6.

Number interior = 6×6 = 36.
Total = 64.

Probability = 36/64 = divide by 4: 9/16.

That is choice D. Choices A, B, C, E correspond to other counts, e.g. 28/64 edge vs interior confusion.
ANSWER 6: D

Problem 7:
We track makes.

After 20 shots at 55%: makes = 0.55 × 20 = 11.

After 5 more shots, total shots = 25, percentage = 56%:
makes = 0.56 × 25 = 14. Note 56% = 56/100 = 14/25, so 14/25 × 25 = 14, a whole number as expected.

Makes in last 5 = 14 − 11 = 3.

So she made 3 of the last 5. Choice C. 5 would give 16/25 = 64%, too high; 1 or 2 would lower the percentage.
ANSWER 7: C

Problem 8:
Book is 760 pages.

Bob: 45 seconds per page. Total = 760 × 45.
Chandra: 30 seconds per page. Total = 760 × 30.

Difference Bob − Chandra = 760×45 − 760×30 = 760×(45−30) = 760×15.

760×15 = 760×10 + 760×5 = 7600 + 3800 = 11,400 seconds.

Choice B. Choice A 7,600 would be 760×10, choice E 22,800 would be forgetting to subtract, etc.
ANSWER 8: B

Problem 9:
Each pitcher holds 600 mL.

Orange juice initially:
Pitcher 1: 1/3 full → (1/3)×600 = 200 mL juice.
Pitcher 2: 2/5 full → (2/5)×600 = 240 mL juice.
Total juice = 200 + 240 = 440 mL.

Water is added to fill each to 600 mL, then both poured together.
Total mixture = 600 + 600 = 1200 mL.

Fraction that is juice = 440 / 1200 = 44/120 = 11/30.

Choice C. Choice E 11/15 is double, forgetting total is two pitchers, etc.
ANSWER 9: C

Problem 10:
Average = sum ÷ count, so sum = average × count.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.

Sum of last three = 270 − 96 = 174.

Average of last three = 174 ÷ 3 = 58.

Choice D. Choice A 55, B 56 etc. would not give total 270.
ANSWER 10: D

Problem 1:
**Restatement:** We need to find how many more minutes it takes Cheenu to travel one mile now as an old man compared to when he was a boy.

**Step-by-step solution:**
1. **Calculate his pace as a boy:**
   - Distance: $15\text{ miles}$
   - Time: $3\text{ hours and } 30\text{ minutes} = 3 \times 60 + 30 = 210\text{ minutes}$
   - Pace: $\frac{210\text{ minutes}}{15\text{ miles}} = 14\text{ minutes per mile}$

2. **Calculate his pace now:**
   - Distance: $10\text{ miles}$
   - Time: $4\text{ hours} = 4 \times 60 = 240\text{ minutes}$
   - Pace: $\frac{240\text{ minutes}}{10\text{ miles}} = 24\text{ minutes per mile}$

3. **Find the difference:**
   - $24 - 14 = 10\text{ minutes}$

It takes him 10 minutes longer to travel a mile now.

ANSWER 1: B

---

Problem 2:
**Restatement:** We need to compute the value of the sum:
$$\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} + \frac{9}{10} + \frac{55}{10}$$

**Step-by-step solution:**
1. Since all terms share the common denominator $10$, we can combine them into a single fraction:
   $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 55}{10}$$

2. Sum the integers from $1$ to $9$:
   $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = \frac{9 \times 10}{2} = 45$$

3. Add $55$:
   $$45 + 55 = 100$$

4. Divide by the denominator:
   $$\frac{100}{10} = 10$$

ANSWER 2: D

---

Problem 3:
**Restatement:** We need to determine which of the given numbers (55, 57, 58, 59, 61) has the smallest prime factor.

**Step-by-step solution:**
1. The smallest prime numbers in increasing order are $2, 3, 5, 7, \dots$
2. Let's find the smallest prime factor of each option:
   - For A ($55$): $55 = 5 \times 11$, so its smallest prime factor is $5$.
   - For B ($57$): $5 + 7 = 12$ is divisible by $3$, so $57 = 3 \times 19$, and its smallest prime factor is $3$.
   - For C ($58$): $58$ is an even number, so it is divisible by $2$ ($58 = 2 \times 29$). Its smallest prime factor is $2$.
   - For D ($59$): $59$ is a prime number, so its smallest prime factor is $59$.
   - For E ($61$): $61$ is a prime number, so its smallest prime factor is $61$.
3. Since $2$ is the smallest prime number possible, $58$ has the smallest prime factor.

ANSWER 3: C

---

Problem 4:
**Restatement:** The sum of two prime numbers is $85$. We need to find their product.

**Step-by-step solution:**
1. The sum of the two primes is $85$, which is an odd number.
2. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is $2$.
4. Therefore, one of the prime numbers must be $2$.
5. The other prime number is:
   $$85 - 2 = 83$$
6. We check if $83$ is prime: since it is not divisible by $2, 3, 5,$ or $7$, $83$ is indeed prime.
7. Now, compute their product:
   $$2 \times 83 = 166$$

ANSWER 4: E

---

Problem 5:
**Restatement:** Lucius counts backward by $7$s starting from $100$. We are asked to find the $10\text{th}$ number in the sequence.

**Step-by-step solution:**
1. The sequence starts at $a_1 = 100$ and decreases by $7$ each step:
   - $a_1 = 100$
   - $a_2 = 100 - 7 = 93$
   - $a_3 = 100 - 2 \times 7 = 86$
2. The $n\text{th}$ term of an arithmetic sequence with initial term $a_1$ and common difference $d = -7$ is given by:
   $$a_n = a_1 + (n - 1)d$$
3. For the $10\text{th}$ number ($n = 10$):
   $$a_{10} = 100 - (10 - 1) \times 7 = 100 - 9 \times 7 = 100 - 63 = 37$$

ANSWER 5: B

---

Problem 6:
**Restatement:** On an $8 \times 8$ checkerboard (64 unit squares total), find the probability that a randomly selected unit square does not touch the outer edge of the board.

**Step-by-step solution:**
1. A standard checkerboard consists of an $8 \times 8$ grid of unit squares, so there are $64$ total unit squares.
2. The squares that touch the outer boundary are located on the outermost ring: rows $1$ and $8$, and columns $1$ and $8$.
3. The squares that *do not* touch the outer edge are the interior squares, which form a square grid of size:
   $$(8 - 2) \times (8 - 2) = 6 \times 6 = 36\text{ squares}$$
4. The probability of choosing one of these inner squares is:
   $$\frac{36}{64} = \frac{9}{16}$$

ANSWER 6: D

---

Problem 7:
**Restatement:** Sally has made $55\%$ of her first $20$ shots. After taking $5$ more shots, her percentage increases to $56\%$. We need to find how many of the last $5$ shots she made.

**Step-by-step solution:**
1. Calculate the number of shots Sally made in her first $20$ shots:
   $$\text{Shots made initially} = 0.55 \times 20 = 11$$
2. After taking $5$ more shots, her total number of shots is:
   $$20 + 5 = 25\text{ shots}$$
3. Her success rate is now $56\%$, so the total number of shots she made is:
   $$\text{Total shots made} = 0.56 \times 25 = 14$$
4. The number of successful shots among the last $5$ is:
   $$14 - 11 = 3$$

ANSWER 7: C

---

Problem 8:
**Restatement:** A novel is $760$ pages long. Bob reads at a rate of $45$ seconds per page and Chandra reads at a rate of $30$ seconds per page. We need to find how many more seconds Bob spends reading the entire novel than Chandra.

**Step-by-step solution:**
1. For each page, Bob takes:
   $$45 - 30 = 15\text{ seconds longer than Chandra}$$
2. For the entire $760$-page novel, the difference in total reading time is:
   $$760 \times 15\text{ seconds}$$
3. Calculate the product:
   $$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11400\text{ seconds}$$

ANSWER 8: B

---

Problem 9:
**Restatement:** Two $600\text{ mL}$ pitchers contain orange juice: one is $\frac{1}{3}$ full and the other is $\frac{2}{5}$ full. Water is added to completely fill each pitcher to $600\text{ mL}$. Both are poured together into a large container. We need to find the fraction of the final mixture that is orange juice.

**Step-by-step solution:**
1. Calculate the volume of orange juice in each pitcher:
   - First pitcher: $\frac{1}{3} \times 600\text{ mL} = 200\text{ mL}$ of orange juice.
   - Second pitch

# Problem 1:
**Find how many minutes longer per mile Cheenu takes now compared to his youth.**

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

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

**Difference:** 24 − 14 = 10 minutes.

*Check wrong answers:* 30 would come from subtracting 210 from 240 (total times, not per-mile); 6 from mis-converting 3.5 hours as 190 minutes. The per-mile calculation gives 10.

ANSWER 1: B

# Problem 2:
**Evaluate** 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10 + 55/10.

All terms share the denominator 10, so add the numerators:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 (sum of 1 through 9)

Then: 45 + 55 = 100

Total = 100/10 = **10**

*Check wrong answers:* 9 would come from forgetting the 55/10 term (45/10 + 45/10); 6.4 ignores the large last term. Only 10 matches the full sum.

ANSWER 2: D

# Problem 3:
**Find which number has the smallest prime factor.** Factor each option:

- **A. 55** = 5 × 11 → smallest prime factor **5**
- **B. 57** = 3 × 19 → smallest prime factor **3**
- **C. 58** = 2 × 29 → smallest prime factor **2**
- **D. 59** is prime → smallest prime factor **59**
- **E. 61** is prime → smallest prime factor **61**

Since 58 is even, its smallest prime factor is 2, which is the smallest possible prime. No other option can beat that.

ANSWER 3: C

# Problem 4:
**Two primes sum to 85; find their product.**

The sum 85 is **odd**. The sum of two odd numbers is even, so two odd primes can never sum to an odd number. Therefore one of the primes must be the only even prime: **2**.

The other prime is 85 − 2 = **83** (which is indeed prime).

Product = 2 × 83 = **166**

*Check wrong answers:* 85 = 5 × 17 uses factors whose sum is 22, not 85; 91 = 7 × 13 (sum 20); 115 = 5 × 23 (sum 28); 133 = 7 × 19 (sum 26). Only 166 works.

ANSWER 4: E

# Problem 5:
**Find the 10th number when counting backward by 7s starting at 100.**

The sequence is 100, 93, 86, 79, ...

The n-th term is given by: 100 − 7(n − 1)

For n = 10:
100 − 7(10 − 1) = 100 − 7(9) = 100 − 63 = **37**

*Check by listing:* 100, 93, 86, 79, 72, 65, 58, 51, 44, **37** ✓

ANSWER 5: B

# Problem 6:
**Find the probability a randomly chosen square does not touch the outer edge.**

A checkerboard is 8 × 8 = 64 unit squares. A square avoids the outer edge only if it is not in the first or last row and not in the first or last column.

Removing the border rows and columns leaves an inner grid of:
(8 − 2) × (8 − 2) = 6 × 6 = **36 squares**

Probability = 36/64 = **9/16**

*Check wrong answers:* 1/16 counts only a 2×2 inner region (subtracting wrongly); 7/16 = 28/64 has no geometric meaning here; 49/64 would be a 7×7 region. The 6×6 interior is correct.

ANSWER 6: D

# Problem 7:
**Find how many of the last 5 shots Sally made.**

**After 20 shots:** 55% made → 0.55 × 20 = **11 shots made**

**After 25 shots:** 56% made → 0.56 × 25 = **14 shots made**

Shots made in the last 5 = 14 − 11 = **3**

*Check:* With 3 makes in the last 5, total = 14/25 = 0.56 = 56% ✓

ANSWER 7: C

# Problem 8:
**Find how many more seconds Bob spends reading the 760-page novel than Chandra.**

**Bob:** 45 seconds per page × 760 pages = 34,200 seconds

**Chandra:** 30 seconds per page × 760 pages = 22,800 seconds

**Difference:** 34,200 − 22,800 = **11,400 seconds**

*(Shortcut: Bob spends 45 − 30 = 15 extra seconds per page, and 15 × 760 = 11,400.)*

*Check wrong answers:* 22,800 is Chandra's total time; 15,200 = 760 × 20 (wrong per-page gap); 7,600 = 760 × 10. The correct gap is 11,400.

ANSWER 8: B

# Problem 9:
**Find the fraction of the final mixture that is orange juice.**

Each pitcher holds 600 mL.

**Pitcher 1:** 1/3 full of juice → juice = 600 × 1/3 = **200 mL**; water added = 400 mL

**Pitcher 2:** 2/5 full of juice → juice = 600 × 2/5 = **240 mL**; water added = 360 mL

**Large container:** total volume = 600 + 600 = 1200 mL; total juice = 200 + 240 = 440 mL

Fraction of juice = 440/1200 = **11/30**

*Check wrong answers:* 11/15 = 440/600 forgets the water; 1/8 or 3/16 are far too small since nearly half of each pitcher was juice. 11/30 is correct.

ANSWER 9: C

# Problem 10:
**Find the average of the last three numbers.**

**Sum of all five numbers:** 54 × 5 = 270

**Sum of the first two numbers:** 48 × 2 = 96

**Sum of the last three numbers:** 270 − 96 = 174

**Average of the last three:** 174 ÷ 3 = **58**

*Check:* If the last three average 58, total = 96 + 174 = 270, and 270 ÷ 5 = 54 ✓

ANSWER 10: D

Problem 1:  
Cheenu’s boy time: \(3\) hours \(30\) minutes \(=210\) minutes.  
Time per mile as a boy:
\[
\frac{210}{15}=14\text{ minutes per mile}
\]
Cheenu’s old-man time: \(4\) hours \(=240\) minutes.  
Time per mile now:
\[
\frac{240}{10}=24\text{ minutes per mile}
\]
The difference is
\[
24-14=10\text{ minutes}
\]
So it now takes \(10\) minutes longer per mile.

ANSWER 1: B

Problem 2:  
The expression is
\[
\frac1{10}+\frac2{10}+\frac3{10}+\frac4{10}+\frac5{10}+\frac6{10}+\frac7{10}+\frac8{10}+\frac9{10}+\frac{55}{10}
\]
Add the numerators:
\[
1+2+3+4+5+6+7+8+9+55=100
\]
So the value is
\[
\frac{100}{10}=10
\]

ANSWER 2: D

Problem 3:  
Find the smallest prime factor of each number:

- \(55=5\cdot 11\), smallest prime factor \(5\)
- \(57=3\cdot 19\), smallest prime factor \(3\)
- \(58=2\cdot 29\), smallest prime factor \(2\)
- \(59\) is prime, smallest prime factor \(59\)
- \(61\) is prime, smallest prime factor \(61\)

The smallest possible prime factor is \(2\), which occurs for \(58\).

ANSWER 3: C

Problem 4:  
The sum of the two primes is \(85\), an odd number.  
Since all primes except \(2\) are odd, the sum of two odd primes would be even. Therefore, one of the primes must be \(2\).  
The other prime is
\[
85-2=83
\]
and \(83\) is prime.  
Their product is
\[
2\cdot 83=166
\]

ANSWER 4: E

Problem 5:  
Lucius counts backward by \(7\):
\[
100,\ 93,\ 86,\ \ldots
\]
The \(n\)th number is
\[
100-7(n-1)
\]
For the 10th number:
\[
100-7(9)=100-63=37
\]

ANSWER 5: B

Problem 6:  
A checkerboard has \(64\) unit squares, arranged \(8\) by \(8\).  
A square that does not touch the outer edge must be one of the interior squares.  
Removing the outer row and column from the top, bottom, left, and right leaves a \(6\) by \(6\) interior region:
\[
6\cdot 6=36
\]
So the probability is
\[
\frac{36}{64}=\frac{9}{16}
\]

ANSWER 6: D

Problem 7:  
After 20 shots, Sally made \(55\%\):
\[
0.55\cdot 20=11
\]
After 25 total shots, she made \(56\%\):
\[
0.56\cdot 25=14
\]
So during the last 5 shots she made
\[
14-11=3
\]

ANSWER 7: C

Problem 8:  
Bob reads a page in \(45\) seconds, and Chandra reads a page in \(30\) seconds.  
For each page, Bob spends
\[
45-30=15
\]
more seconds than Chandra.  
For the whole \(760\)-page book, the difference is
\[
760\cdot 15=11400
\]

ANSWER 8: B

Problem 9:  
Each pitcher has capacity \(600\) mL.

First pitcher:
\[
\frac13\cdot 600=200\text{ mL of orange juice}
\]

Second pitcher:
\[
\frac25\cdot 600=240\text{ mL of orange juice}
\]

Total orange juice:
\[
200+240=440
\]

Total mixture:
\[
600+600=1200
\]

Fraction that is orange juice:
\[
\frac{440}{1200}=\frac{11}{30}
\]

ANSWER 9: C

Problem 10:  
The average of five numbers is \(54\), so their total sum is
\[
5\cdot 54=270
\]
The average of the first two numbers is \(48\), so their sum is
\[
2\cdot 48=96
\]
The sum of the last three numbers is
\[
270-96=174
\]
Their average is
\[
\frac{174}{3}=58
\]

ANSWER 10: D

# Problem 1:
**Asked:** How many more minutes per mile does he need now vs. as a boy?

**As a boy:** 15 miles in 3 h 30 min = 210 minutes.
Time per mile = 210 ÷ 15 = **14 minutes**.

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

**Difference:** 24 − 14 = 10 minutes.

Ruling out traps: 30 comes from subtracting total times in some wrong way (240 − 210 = 30 is the difference in *total* time, not per mile); 6 would be 24 − 18 (using 18 min/mile, a miscomputation of 210/15).

ANSWER 1: B

# Problem 2:
**Asked:** Evaluate 1/10 + 2/10 + ... + 9/10 + 55/10.

All have denominator 10, so add numerators:
1+2+3+4+5+6+7+8+9 = 45, then 45 + 55 = 100.

Sum = 100/10 = 10.

The trap 9 or 11 would come from mis-adding 45 + 55; 4½ is just (1+…+9)/10 without the 55/10 term.

ANSWER 2: D

# Problem 3:
**Asked:** Which number has the smallest prime factor?

- 55 = 5 · 11 → smallest prime factor 5
- 57 = 3 · 19 → smallest prime factor 3
- 58 = 2 · 29 → smallest prime factor **2**
- 59 is prime → smallest prime factor 59
- 61 is prime → smallest prime factor 61

The smallest possible prime factor is 2, and only the even number 58 has it. (Choosing 55 or 57 confuses "smallest number" ideas; 59 and 61 have huge prime factors since they're prime.)

ANSWER 3: C

# Problem 4:
**Asked:** Two primes sum to 85; find their product.

85 is odd. A sum of two numbers is odd only when one is even and one is odd. The only even prime is 2, so one prime must be 2.

Then the other is 85 − 2 = 83, which is prime (not divisible by 2, 3, 5, 7 — and 9² = 81 < 83 < 100, so we only check up to 9).

Product = 2 · 83 = 166.

(85, 91, 115, 133 are products of odd-prime pairs that don't sum to 85.)

ANSWER 4: E

# Problem 5:
**Asked:** The 10th term of 100, 93, 86, ... (subtracting 7 each time).

The nth number is 100 − 7(n − 1).
For n = 10: 100 − 7(9) = 100 − 63 = 37.

Check by listing: 100, 93, 86, 79, 72, 65, 58, 51, 44, **37**. (44 is the 9th number — the common off-by-one trap; 30 would be the 11th.)

ANSWER 5: B

# Problem 6:
**Asked:** Probability a random unit square on an 8×8 board does **not** touch the outer edge.

Squares touching the edge are those in row 1 or 8, or column 1 or 8. Removing the border leaves the inner 6×6 block:
6 × 6 = 36 interior squares.

Probability = 36/64 = 9/16.

Traps: 7/16 = 28/64 is the border-count fraction miscount; 49/64 would come from a 7×7 inner square (wrong—removing a border takes off 2 from each dimension, not 1).

ANSWER 6: D

# Problem 7:
**Asked:** How many of the last 5 shots did Sally make?

After 20 shots: 55% of 20 = 0.55 · 20 = 11 shots made.
After 25 shots: 56% of 25 = 0.56 · 25 = 14 shots made.

Made in last 5 shots = 14 − 11 = 3.

(Answer 1 or 2 would come from thinking a 1% rise needs very few makes; you must compare actual counts.)

ANSWER 7: C

# Problem 8:
**Asked:** How many more seconds does Bob spend than Chandra reading all 760 pages?

Bob: 760 × 45 = 34,200 seconds.
Chandra: 760 × 30 = 22,800 seconds.

Difference: 34,200 − 22,800 = 11,400 seconds.

Shortcut: 760 × (45 − 30) = 760 × 15 = 11,400.
(22,800 is Chandra's *total* time, not the difference — a common trap.)

ANSWER 8: B

# Problem 9:
**Asked:** Fraction of orange juice in the combined 1200 mL.

Pitcher 1 juice: (1/3)(600) = 200 mL.
Pitcher 2 juice: (2/5)(600) = 240 mL.

Each pitcher is filled to 600 mL with water, so total liquid = 600 + 600 = 1200 mL.
Total juice = 200 + 240 = 440 mL.

Fraction = 440/1200 = 11/30.

Note: averaging 1/3 and 2/5 gives (1/3 + 2/5)/2 = (11/15)/2 = 11/30 — same answer. The trap 11/15 is the sum of the fractions (forgetting to average/divide by the total 1200); 11/19 wrongly compares juice to water.

ANSWER 9: C

# Problem 10:
**Asked:** Average of the last three of five numbers.

Sum of all five = 5 × 54 = 270.
Sum of first two = 2 × 48 = 96.
Sum of last three = 270 − 96 = 174.

Average = 174 ÷ 3 = 58.

(Just saying "54 + 6 = 60" or averaging averages gives wrong results; you must use sums.)

ANSWER 10: D

Problem 1:

We are asked how many **more minutes per mile** Cheenu takes now than when he was a boy.

When he was a boy:

- \(3\) hours \(30\) minutes \(= 210\) minutes.
- He ran \(15\) miles in \(210\) minutes.

So his time per mile was

\[
210 \div 15 = 14 \text{ minutes per mile}.
\]

Now:

- He walks \(10\) miles in \(4\) hours \(=240\) minutes.

So his time per mile now is

\[
240 \div 10=24 \text{ minutes per mile}.
\]

The increase is

\[
24-14=10.
\]

Choice A, 6, would not account for finding the time per mile. Choices such as 15 or 18 come from mixing up the total times or distances.

ANSWER 1: B

Problem 2:

The expression is

\[
\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\cdots+\frac{9}{10}+\frac{55}{10}.
\]

Since all terms have denominator \(10\), add the numerators:

\[
1+2+3+4+5+6+7+8+9=45.
\]

Thus,

\[
\frac{45}{10}+\frac{55}{10}
=\frac{100}{10}
=10.
\]

Choice C, 9, would result from using only \(1+2+\cdots+9=45\) incorrectly. Choice E, 11, is too large.

ANSWER 2: D

Problem 3:

We need the number with the **smallest prime factor**.

Check each number:

- \(55=5\cdot 11\), so its smallest prime factor is \(5\).
- \(57=3\cdot 19\), so its smallest prime factor is \(3\).
- \(58=2\cdot 29\), so its smallest prime factor is \(2\).
- \(59\) is prime, so its smallest prime factor is \(59\).
- \(61\) is prime, so its smallest prime factor is \(61\).

The smallest possible prime factor here is \(2\), and it belongs to \(58\).

A tempting mistake is to choose 59 because it is the smallest prime number listed, but the question asks for the number having the smallest prime **factor**.

ANSWER 3: C

Problem 4:

The sum of two prime numbers is \(85\), which is odd.

All prime numbers except \(2\) are odd. The sum of two odd numbers is even, so one of the primes must be \(2\).

The other prime is

\[
85-2=83.
\]

Since \(83\) is prime, the two primes are \(2\) and \(83\).

Their product is

\[
2\cdot 83=166.
\]

Choices such as 91 or 133 are products of other primes, but their prime factors do not add to 85.

ANSWER 4: E

Problem 5:

Lucius counts backward by 7s:

\[
100,\ 93,\ 86,\ldots
\]

Each new number is \(7\) less than the previous one.

To get from the 1st number to the 10th number, we subtract \(7\) a total of \(9\) times:

\[
100-9(7)=100-63=37.
\]

So his 10th number is \(37\).

A common error is subtracting \(7\) ten times, but the first number, 100, is already counted as the 1st number.

ANSWER 5: B

Problem 6:

A checkerboard with 64 unit squares is an \(8\times 8\) board.

Squares that do **not** touch the outer edge must be completely inside the border. Removing the outer row and column from each side leaves a

\[
6\times 6
\]

interior region.

Thus the number of squares not touching the edge is

\[
6\cdot 6=36.
\]

There are \(64\) total squares, so the probability is

\[
\frac{36}{64}=\frac{9}{16}.
\]

Choice E, \(\frac{49}{64}\), would come from incorrectly thinking there are \(7\times7\) interior squares. The outer border must be removed from both opposite sides.

ANSWER 6: D

Problem 7:

After 20 shots, Sally has made \(55\%\) of them:

\[
0.55\cdot 20=11.
\]

So she has made 11 shots after her first 20 attempts.

After 5 more shots, she has taken

\[
20+5=25
\]

shots total. Her new percentage is \(56\%\), so she has made

\[
0.56\cdot 25=14
\]

shots total.

Therefore, among the last 5 shots, she made

\[
14-11=3.
\]

A wrong answer such as 4 or 5 would make her final total higher than 14 made shots.

ANSWER 7: C

Problem 8:

Bob reads each page in 45 seconds, while Chandra reads each page in 30 seconds.

For each page, Bob takes

\[
45-30=15
\]

more seconds than Chandra.

There are 760 pages, so Bob spends

\[
760\cdot 15=11,\!400
\]

more seconds reading.

Choice D, 15,200, would come from multiplying by 20 rather than the correct per-page difference of 15 seconds.

ANSWER 8: B

Problem 9:

Each pitcher holds 600 mL.

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

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

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

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

Thus the total orange juice is

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

After both pitchers are filled completely and poured together, the total mixture is

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

Therefore, the fraction that is orange juice is

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

Choice D, \(\frac{11}{19}\), incorrectly compares the orange juice to only the added water rather than to the whole mixture.

ANSWER 9: C

Problem 10:

The average of five numbers is 54, so their total is

\[
5\cdot 54=270.
\]

The average of the first two numbers is 48, so their total is

\[
2\cdot 48=96.
\]

Therefore, the sum of the last three numbers is

\[
270-96=174.
\]

Their average is

\[
174\div 3=58.
\]

A common mistake is to subtract the averages, \(54-48\), but averages cannot be subtracted directly because they represent groups of different sizes.

ANSWER 10: D