🔬 Model Lab

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B

I'll solve each problem step-by-step.

---

**Problem 1:**

**What is being asked:** What fraction of the pizza did Peter eat?

**Setup:** 
- Pizza has 12 slices total
- Peter ate 1 whole slice
- Peter shared another slice equally with Paul, so Peter got 1/2 of that slice
- Total Peter ate: 1 + 1/2 = 3/2 slices

**Computation:**
Peter ate 3/2 slices out of 12 total slices.
Fraction = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8

**Check answer choices:** This matches choice C.

ANSWER 1: C

---

**Problem 2:**

**What is being asked:** How much could Karl have saved by waiting for the 20%-off sale?

**Setup:**
- 5 folders at $2.50 each
- Original cost: 5 × $2.50 = $12.50
- Sale is 20% off

**Computation:**
20% of $12.50 = 0.20 × $12.50 = $2.50

**Check answer choices:** This matches choice C.

ANSWER 2: C

---

**Problem 3:**

**What is being asked:** How many clerts are in a right angle?

**Setup:**
- 500 clerts = 360° (full circle)
- Right angle = 90°
- Need to find how many clerts = 90°

**Computation:**
If 500 clerts = 360°, then:
clerts per degree = 500/360 = 25/18
clerts in 90° = 90 × (500/360) = 90 × 500/360 = 45000/360 = 125

**Check answer choices:** This matches choice C.

ANSWER 3: C

---

**Problem 4:**

**What is being asked:** For x = 7, which expression is smallest?

**Setup:** Substitute x = 7 into each choice:

**Computation:**
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x-1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

The smallest value is 6/8 = 0.75

**Check answer choices:** This matches choice B.

ANSWER 4: B

---

**Problem 5:**

**What is being asked:** What is the sum of the four corner numbers on an 8×8 checkerboard with numbers 1-64?

**Setup:**
- Numbers 1-8 in row 1
- Numbers 9-16 in row 2
- And so on...
- Row 8 has numbers 57-64

**Computation:**
- Top-left corner: 1
- Top-right corner: 8
- Bottom-left corner: 57 (start of row 8)
- Bottom-right corner: 64

Sum = 1 + 8 + 57 + 64 = 130

**Check answer choices:** This matches choice A.

ANSWER 5: A

---

**Problem 6:**

**What is being asked:** How many digits are in 4^5 · 5^10?

**Setup:**
4^5 · 5^10 = (2^2)^5 · 5^10 = 2^10 · 5^10 = (2 · 5)^10 = 10^10

**Computation:**
10^10 = 10,000,000,000 (1 followed by 10 zeros)

Number of digits = 11

**Check answer choices:** This matches choice D.

ANSWER 6: D

---

**Problem 7:**

**What is being asked:** If 2 more friends join, how many cards per player?

**Setup:**
- Current game: Annika + 3 friends = 4 players, each gets 15 cards
- Total cards: 4 × 15 = 60 cards
- New game: 4 + 2 = 6 players

**Computation:**
Cards per player = 60 ÷ 6 = 10

**Check answer choices:** This matches choice C.

ANSWER 7: C

---

**Problem 8:**

**What is being asked:** Calculate (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

**Setup:** Add the two groups

**Computation:**
First group: 1 + 11 + 21 + 31 + 41 = 105
Second group: 9 + 19 + 29 + 39 + 49 = 145
Total: 105 + 145 = 250

**Check answer choices:** This matches choice E.

ANSWER 8: E

---

**Problem 9:**

**What is being asked:** What is the largest number Ahn can get by computing 2(200 - n) where n is a two-digit integer?

**Setup:**
Result = 2(200 - n)
To maximize this, we need to minimize n
Smallest two-digit integer is 10

**Computation:**
2(200 - 10) = 2(190) = 380

**Check answer choices:** This matches choice D.

ANSWER 9: D

---

**Problem 10:**

**What is being asked:** How old is Caitlin?

**Setup:**
- Aunt Anna is 42 years old
- Brianna is half as old as Aunt Anna: Brianna = 42/2 = 21
- Caitlin is 5 years younger than Brianna: Caitlin = 21 - 5 = 16

**Computation:**
Caitlin = 16 years old

**Check answer choices:** This matches choice B.

ANSWER 10: B

Problem 1:

Peter ate:
- 1 whole slice, plus
- half of another slice.

Since the pizza has 12 slices, one slice is \( \frac{1}{12} \) of the pizza.

Half of a slice is:
\[
\frac{1}{2}\cdot\frac{1}{12}=\frac{1}{24}
\]

So Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}
\]

So the fraction of the pizza Peter ate is \( \frac{1}{8} \).

ANSWER 1: C

Problem 2:

Karl bought 5 folders at \$2.50 each, so the original total cost was:
\[
5 \times 2.50 = 12.50
\]

A 20% discount means he would save 20% of \$12.50:
\[
0.20 \times 12.50 = 2.50
\]

So Karl could have saved \$2.50 by waiting.

ANSWER 2: C

Problem 3:

A full circle is 500 clerts. A right angle is one-fourth of a full circle, so:
\[
\frac{500}{4} = 125
\]

So there are 125 clerts in a right angle.

ANSWER 3: C

Problem 4:

We substitute \(x=7\) into each expression.

- A: \(6x = 6(7)=42\)
- B: \(6x+1 = 42+1=43\)
- C: \(6x-1 = 42-1=41\)
- D: \(x^6 = 7^6\), which is much larger than 42
- E: \(x+1^6\) is interpreted as \(x+1=8\) or possibly \(x+(1^6)=8\), either way it is greater than 41? Actually \(1^6=1\), so this is \(7+1=8\)

Now compare the values:
\[
42,\ 43,\ 41,\ 7^6,\ 8
\]
The smallest is \(8\), which is choice E.

ANSWER 4: E

Problem 5:

The numbers 1 through 64 are placed in rows of 8:

- First row: 1 to 8
- Second row: 9 to 16
- ...
- Eighth row: 57 to 64

So the four corners are:
- top left: 1
- top right: 8
- bottom left: 57
- bottom right: 64

Add them:
\[
1+8+57+64 = 130
\]

ANSWER 5: A

Problem 6:

The expression is
\[
4^5 \cdot 5^{10}
\]

Compute:
\[
4^5 = (2^2)^5 = 2^{10}
\]
So
\[
4^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}
\]

The number \(10^{10}\) is 1 followed by 10 zeros, so it has:
\[
11
\]
digits.

ANSWER 6: D

Problem 7:

With Annika and 3 friends, there are:
\[
4 \text{ players}
\]
Each gets 15 cards, so total cards are:
\[
4 \times 15 = 60
\]

Now 2 more friends join, so there are:
\[
6 \text{ players}
\]

Split 60 cards evenly among 6 players:
\[
60 \div 6 = 10
\]

So each player gets 10 cards.

ANSWER 7: C

Problem 8:

Add the first group:
\[
1+11+21+31+41
\]
These are 5 numbers in an arithmetic pattern. Pair them:
\[
1+41=42,\quad 11+31=42
\]
So:
\[
1+11+21+31+41 = 42+42+21 = 105
\]

Add the second group:
\[
9+19+29+39+49
\]
Pair them:
\[
9+49=58,\quad 19+39=58
\]
So:
\[
9+19+29+39+49 = 58+58+29 = 145
\]

Total:
\[
105+145=250
\]

ANSWER 8: E

Problem 9:

Ahn chooses a two-digit integer. The largest two-digit integer is 99.

He subtracts it from 200:
\[
200-99=101
\]

Then he doubles the result:
\[
2 \times 101 = 202
\]

So the largest possible number is 202.

ANSWER 9: B

Problem 10:

Aunt Anna is 42.

Brianna is half as old as Aunt Anna:
\[
42 \div 2 = 21
\]

Caitlin is 5 years younger than Brianna:
\[
21-5=16
\]

So Caitlin is 16 years old.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one slice and shared another slice equally with his brother Paul.  
- One slice means Peter ate \(1\) out of \(12\).  
- The shared slice: if they share one slice equally, Peter gets \(\frac{1}{2}\) slice, i.e. \(\frac{1/2}{12}=\frac{1}{24}\) of the pizza.  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]
This matches choice C.

ANSWER 1: C  

---

Problem 2:  
Karl bought five folders at \(\$2.50\) each. So the original cost was:
\[
5 \times 2.50 = 12.50.
\]
A 20%-off sale would reduce the price by:
\[
0.20 \times 12.50 = 2.50.
\]
So he could have saved \(\$2.50\). Choice C.

ANSWER 2: C  

---

Problem 3:  
A full circle has 500 clerts. A right angle is \(\frac{1}{4}\) of a circle.  
\[
500 \div 4 = 125.
\]
So the right angle is 125 clerts. Choice C.

ANSWER 3: C  

---

Problem 4:  
For \(x=7\), find the smallest among:
A. \(6x\)  
B. \(6x+1\)  
C. \(6x-1\)  
D. \(x^6\)  
E. \(\frac{x+1}{6}\)  (since the formatting “\(x +1 6\)” corresponds to \(\frac{x+1}{6}\))

Compute each:
- \(6x = 6\cdot 7 = 42\)
- \(6x+1 = 42+1 = 43\)
- \(6x-1 = 42-1 = 41\)
- \(x^6 = 7^6 = 117649\) (very large)
- \(\frac{x+1}{6}=\frac{8}{6}=\frac{4}{3}\approx 1.33\)

The smallest is \(\frac{4}{3}\), which is choice E.

ANSWER 4: E  

---

Problem 5:  
We place numbers 1 through 64 in order across rows of an \(8\times 8\) checkerboard.

Corners:
- Top-left is \(1\).
- Top-right is \(8\).
- Bottom-left is \(57\) (since row 8 starts at \(8\cdot 7+1=57\)).
- Bottom-right is \(64\).

Sum of corners:
\[
1+8+57+64 =  (1+8) + (57+64) = 9 + 121 = 130.
\]
Choice A.

ANSWER 5: A  

---

Problem 6:  
“How many digits are in the product \(4^5 \cdot 5^{10}\)?”  
Compute exponent rule:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10}\cdot 5^{10} = (2\cdot 5)^{10} = 10^{10}.
\]
\(10^{10}\) is \(1\) followed by 10 zeros, so it has \(11\) digits.

Choice D.

ANSWER 6: D  

---

Problem 7:  
Initially, Annika and 3 friends = 4 players total. Each gets 15 cards, so total cards:
\[
4 \times 15 = 60.
\]
Next game: 2 more friends join. Total players:
\[
4 + 2 = 6.
\]
Each gets:
\[
60 \div 6 = 10.
\]
Choice C.

ANSWER 7: C  

---

Problem 8:  
\[
(1+11+21+31+41) + (9+19+29+39+49).
\]
Notice each group is an arithmetic sequence with common difference 10.

First group:
\[
1,11,21,31,41
\]
Sum = \(5\) terms:
\[
\text{avg}=\frac{1+41}{2}=21 \quad\Rightarrow\quad 5\cdot 21=105.
\]

Second group:
\[
9,19,29,39,49
\]
Average:
\[
\frac{9+49}{2}=29 \quad\Rightarrow\quad 5\cdot 29=145.
\]

Total:
\[
105+145=250.
\]
Choice E.

ANSWER 8: E  

---

Problem 9:  
Ahn chooses a two-digit integer \(n\) (so \(10\le n \le 99\)).  
He subtracts it from 200: \(200-n\), then doubles:
\[
2(200-n)=400-2n.
\]
To make this largest, we need \(2n\) smallest, so choose the smallest two-digit integer: \(n=10\).
\[
400-2(10)=400-20=380.
\]
Choice D.

ANSWER 9: D  

---

Problem 10:  
Aunt Anna is 42.  
Brianna is half as old as Aunt Anna, so:
\[
\text{Brianna}=\frac{1}{2}\cdot 42=21.
\]
Caitlin is 5 years younger than Brianna:
\[
\text{Caitlin}=21-5=16.
\]
Choice B.

ANSWER 10: B

Problem 1:
Peter ate one whole slice, plus half of another slice. Since the pizza has 12 slices, one slice is $1/12$ of the pizza, and half a slice is $1/24$ of the pizza.
Total fraction eaten = $1/12 + 1/24 = 2/24 + 1/24 = 3/24$.
Simplifying the fraction $3/24 = 1/8$.
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, for a total cost of $5 \times 2.50 = $12.50$.
A 20% discount on the total cost would save him $20\%$ of $12.50$.
$0.20 \times 12.50 = 2.50$.
Alternatively, $20\%$ of one folder is $0.20 \times 2.50 = 0.50$. Since there are 5 folders, the total savings is $5 \times 0.50 = 2.50$.
ANSWER 2: C

Problem 3:
A full circle is 500 clerts. A right angle is $1/4$ of a full circle.
Number of clerts in a right angle = $500 / 4 = 125$.
ANSWER 3: C

Problem 4:
Given $x = 7$:
A. $6/x = 6/7 \approx 0.857$
B. $6/(x+1) = 6/8 = 0.75$
C. $6/(x-1) = 6/6 = 1$
D. $x/6 = 7/6 \approx 1.167$
E. $(x+1)/6 = 8/6 \approx 1.333$
Comparing the values, $0.75$ is the smallest.
ANSWER 4: B

Problem 5:
The checkerboard is 8x8. The numbers are:
Row 1: 1, 2, ..., 8 (Corners: 1, 8)
Row 8: 57, 58, ..., 64 (Corners: 57, 64)
Sum of corners = $1 + 8 + 57 + 64 = 130$.
ANSWER 5: A

Problem 6:
The product is $4^5 \cdot 5^{10}$.
We can rewrite $4^5$ as $(2^2)^5 = 2^{10}$.
The expression becomes $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$.
$10^{10}$ is a 1 followed by 10 zeros, which is an 11-digit number.
ANSWER 6: D

Problem 7:
Total cards = (Annika + 3 friends) $\times 15$ cards/player = $4 \times 15 = 60$ cards.
If 2 more friends join, there are $4 + 2 = 6$ players.
Cards per player = $60 / 6 = 10$.
ANSWER 7: C

Problem 8:
Group the terms: $(1+49) + (11+39) + (21+29) + 31 + 41 + 9 + 19$.
Actually, simpler: $(1+11+21+31+41) = 105$.
$(9+19+29+39+49) = 145$.
$105 + 145 = 250$.
ANSWER 8: E

Problem 9:
Let the two-digit integer be $n$. We want to maximize $2(200 - n)$.
To maximize this, we must minimize $n$. The smallest two-digit integer is $n = 10$.
Result = $2(200 - 10) = 2(190) = 380$.
ANSWER 9: D

Problem 10:
Aunt Anna = 42.
Brianna = $42 / 2 = 21$.
Caitlin = $21 - 5 = 16$.
ANSWER 10: B

Problem 1:
Peter ate one full slice out of 12, which is the fraction \( \frac{1}{12} \). He also shared one additional slice equally with Paul, so he ate half of that slice: \( \frac{1}{2} \times \frac{1}{12} = \frac{1}{24} \). Adding these amounts gives the total fraction eaten:
\[
\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}.
\]
This matches choice C (the other choices are either too small or too large).

ANSWER 1: C

Problem 2:
The total cost without the discount is \( 5 \times 2.50 = 12.50 \). A 20% discount on this amount is
\[
0.20 \times 12.50 = 2.50.
\]
Thus Karl would have saved exactly $2.50 by waiting for the sale, which matches choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts, so a right angle (one-fourth of a circle) contains
\[
\frac{1}{4} \times 500 = 125
\]
clerts. This matches choice C.

ANSWER 3: C

Problem 4:
Substitute \( x = 7 \) into each expression (interpreting the ambiguous notation as the natural algebraic forms consistent with an AMC 8 problem: \( 6x \), \( 6x+1 \), \( 6x-1 \), \( x/6 \), and \( x + 1/6 \)):
- A: \( 6 \times 7 = 42 \)
- B: \( 6 \times 7 + 1 = 43 \)
- C: \( 6 \times 7 - 1 = 41 \)
- D: \( 7/6 \approx 1.166 \)
- E: \( 7 + 1/6 \approx 7.166 \)
The numerically smallest value is \( 7/6 \), which is choice D.

ANSWER 4: D

Problem 5:
The numbers 1 through 64 are placed in row-major order on an 8-by-8 board. The four corner entries are therefore
- top-left: 1,
- top-right: 8,
- bottom-left: 57,
- bottom-right: 64.
Their sum is
\[
1 + 8 + 57 + 64 = 130,
\]
which matches choice A.

ANSWER 5: A

Problem 6:
Rewrite the product using powers of 2 and 5:
\[
4^5 = (2^2)^5 = 2^{10}, \qquad 5^{10} = 5^{10}.
\]
Multiplying yields
\[
2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}.
\]
The number \( 10^{10} \) is 1 followed by ten zeros and therefore contains 11 digits, matching choice D.

ANSWER 6: D

Problem 7:
Four players each receive 15 cards, so the deck contains \( 4 \times 15 = 60 \) cards. When two more players join, the same 60 cards are dealt to six players:
\[
\frac{60}{6} = 10
\]
cards each. This matches choice C.

ANSWER 7: C

Problem 8:
The first sum is an arithmetic series of five terms whose average is \( (1+41)/2 = 21 \), so the sum equals \( 5 \times 21 = 105 \). The second sum is an arithmetic series whose average is \( (9+49)/2 = 29 \), so the sum equals \( 5 \times 29 = 145 \). Adding these results gives
\[
105 + 145 = 250,
\]
which matches choice E.

ANSWER 8: E

Problem 9:
Let \( n \) be any two-digit integer. The expression formed is
\[
2(200 - n) = 400 - 2n.
\]
Because the coefficient of \( n \) is negative, the expression is maximized when \( n \) is minimized. The smallest two-digit integer is 10, and
\[
400 - 2 \times 10 = 380.
\]
This matches choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42, so Brianna is half that age:
\[
\frac{42}{2} = 21.
\]
Caitlin is five years younger than Brianna:
\[
21 - 5 = 16.
\]
This matches choice B.

ANSWER 10: B

Problem 1:
We are given that Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. We need to find the fraction of the pizza that Peter ate.
First, let's determine the total number of slices Peter ate. He ate one slice and half of another, so he ate 1 + 1/2 = 3/2 slices.
To find the fraction of the pizza Peter ate, we need to divide the number of slices he ate by the total number of slices.
The fraction of the pizza Peter ate is (3/2) / 12 = 3/24 = 1/8.
Let's examine the choices:
A. 1/24 is too small because Peter ate more than 1 slice.
B. 1/12 is also too small for the same reason.
C. 1/8 matches our calculation.
D. 1/6 is too large because Peter ate less than 2 slices.
E. 1/4 is too large for the same reason.
Therefore, the correct fraction is 1/8.

ANSWER 1: C

Problem 2:
Karl bought five folders from Pay-A-Lot at a cost of $2.50 each. We need to find out how much Karl could have saved on the purchase by waiting a day when Pay-A-Lot had a 20%-off sale.
First, let's calculate the original cost of the folders: 5 * $2.50 = $12.50.
The discount on the sale day is 20% of $12.50, which is 0.20 * $12.50 = $2.50.
So, Karl could have saved $2.50 by waiting a day.
Let's examine the choices:
A. $1.00 is too low.
B. $2.00 is less than the calculated savings.
C. $2.50 matches our calculation.
D. $2.75 is more than the calculated savings.
E. $5.00 is too high.
Therefore, the amount Karl could have saved is $2.50.

ANSWER 2: C

Problem 3:
Martians measure angles in clerts, and there are 500 clerts in a full circle. We need to find out how many clerts are there in a right angle.
A full circle is 360 degrees, and a right angle is 90 degrees. 
The fraction of a full circle that a right angle represents is 90/360 = 1/4.
Since there are 500 clerts in a full circle, the number of clerts in a right angle is 1/4 * 500 = 125.
Let's examine the choices:
A. 90 is the degree measure of a right angle, not clerts.
B. 100 is less than our calculated answer.
C. 125 matches our calculation.
D. 180 is the degree measure of a straight angle, not clerts.
E. 250 is more than our calculated answer.
Therefore, the number of clerts in a right angle is 125.

ANSWER 3: C

Problem 4:
We are given x = 7 and need to determine which of the given expressions is the smallest.
Let's evaluate each expression:
A. 6/x = 6/7 ≈ 0.857
B. 6/x + 1 = 6/7 + 1 = 0.857 + 1 = 1.857
C. 6/x - 1 = 6/7 - 1 = 0.857 - 1 = -0.143
D. x/6 = 7/6 ≈ 1.167
E. (x + 1)/6 = (7 + 1)/6 = 8/6 ≈ 1.333
Comparing the results:
A. 0.857
B. 1.857
C. -0.143
D. 1.167
E. 1.333
The smallest value is -0.143, which corresponds to choice C.

ANSWER 4: C

Problem 5:
The 64 whole numbers from 1 through 64 are written on an 8 by 8 checkerboard. We need to find the sum of the numbers in the four corners.
The numbers in the four corners are the first number in the first row (1), the last number in the first row (8), the first number in the last row (57), and the last number in the last row (64).
Let's add these numbers: 1 + 8 + 57 + 64 = 130.
Let's examine the choices:
A. 130 matches our calculation.
B. 131 is more than our calculated answer.
C. 132 is more than our calculated answer.
D. 133 is more than our calculated answer.
E. 134 is more than our calculated answer.
Therefore, the sum of the numbers in the four corners is 130.

ANSWER 5: A

Problem 6:
We need to find the number of digits in the product 4^5 * 5^10.
First, let's simplify the expression: 4^5 * 5^10 = (2^2)^5 * 5^10 = 2^10 * 5^10 = (2 * 5)^10 = 10^10.
10^10 is a 1 followed by 10 zeros, so it has 11 digits.
Let's examine the choices:
A. 8 is too few digits.
B. 9 is too few digits.
C. 10 is too few digits.
D. 11 matches our calculation.
E. 12 is too many digits.
Therefore, the number of digits in the product is 11.

ANSWER 6: D

Problem 7:
When Annika and 3 of her friends play Buffalo Shuffle-o, each player is dealt 15 cards. We need to find out how many cards will be dealt to each player if 2 more friends join the next game.
First, let's determine the total number of cards: 4 players * 15 cards/player = 60 cards.
In the next game, there will be 4 + 2 = 6 players.
To find the number of cards dealt to each player, we divide the total number of cards by the number of players: 60 cards / 6 players = 10 cards/player.
Let's examine the choices:
A. 8 is too few cards.
B. 9 is too few cards.
C. 10 matches our calculation.
D. 11 is too many cards.
E. 12 is too many cards.
Therefore, the number of cards dealt to each player will be 10.

ANSWER 7: C

Problem 8:
We need to evaluate the expression (1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49).
First, let's calculate the sum of the first sequence: 1 + 11 + 21 + 31 + 41 = 105.
Then, calculate the sum of the second sequence: 9 + 19 + 29 + 39 + 49 = 145.
Now, add the two sums: 105 + 145 = 250.
Let's examine the choices:
A. 150 is too low.
B. 199 is too low.
C. 200 is too low.
D. 249 is too low.
E. 250 matches our calculation.
Therefore, the sum of the given expression is 250.

ANSWER 8: E

Problem 9:
Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result. We need to find the largest number Ahn can get.
To maximize the result, we need to maximize the difference between 200 and the chosen two-digit integer.
The smallest two-digit integer is 10, so the largest difference is 200 - 10 = 190.
Doubling this difference gives 2 * 190 = 380.
Let's examine the choices:
A. 200 is too low.
B. 202 is too low.
C. 220 is too low.
D. 380 matches our calculation.
E. 398 is too high because it would require a negative or non-integer result for the chosen number.
Therefore, the largest number Ahn can get is 380.

ANSWER 9: D

Problem 10:
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. We need to find Caitlin's age.
First, let's determine Brianna's age: 1/2 * Aunt Anna's age = 1/2 * 42 = 21.
Since Caitlin is 5 years younger than Brianna, Caitl

Problem 1:
Peter ate 1 whole slice and shared another slice equally with his brother, so he ate half of that second slice.  
Total slices Peter ate = 1 + 1/2 = 3/2 slices.  
Whole pizza = 12 slices.  
Fraction of pizza = (3/2) ÷ 12 = (3/2) × (1/12) = 3/24 = 1/8.  
ANSWER 1: C

Problem 2:
Cost for 5 folders at $2.50 each = 5 × 2.50 = $12.50.  
With 20% off, each folder costs 0.8 × 2.50 = $2.00.  
Total sale cost = 5 × 2.00 = $10.00.  
Savings = 12.50 − 10.00 = $2.50.  
ANSWER 2: C

Problem 3:
A full circle = 360° = 500 clerts.  
A right angle = 90°.  
Clerts in a right angle = (90/360) × 500 = (1/4) × 500 = 125 clerts.  
ANSWER 3: C

Problem 4:
Substitute x = 7 into each choice:  
A) 6/x = 6/7 ≈ 0.857  
B) 6/(x+1) = 6/8 = 0.75  
C) 6/(x−1) = 6/6 = 1  
D) x/6 = 7/6 ≈ 1.167  
E) (x+1)/6 = 8/6 ≈ 1.333  
The smallest value is 0.75, choice B.  
ANSWER 4: B

Problem 5:
The checkerboard is 8×8, numbers 1 to 64 written row by row.  
Top-left corner: 1  
Top-right corner: 8  
Bottom-left corner: first number of row 8 = 7×8 + 1 = 57  
Bottom-right corner: 64  
Sum = 1 + 8 + 57 + 64 = 130.  
ANSWER 5: A

Problem 6:
4⁵ = (2²)⁵ = 2¹⁰.  
So 4⁵ × 5¹⁰ = 2¹⁰ × 5¹⁰ = (2×5)¹⁰ = 10¹⁰.  
10¹⁰ is a 1 followed by 10 zeros → 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends). Each gets 15 cards ⇒ total cards = 4 × 15 = 60.  
Next game, 2 more friends join ⇒ total players = 6.  
Cards per player = 60 ÷ 6 = 10.  
ANSWER 7: C

Problem 8:
(1+11+21+31+41) + (9+19+29+39+49)  
Pair each term: 1+49=50, 11+39=50, 21+29=50, 31+19=50, 41+9=50.  
Five pairs of 50 = 5 × 50 = 250.  
ANSWER 8: E

Problem 9:
Ahn’s number: 2 × (200 − n), where n is a two‑digit integer (10 to 99).  
To maximize the result, choose the smallest n = 10.  
2 × (200 − 10) = 2 × 190 = 380.  
ANSWER 9: D

Problem 10:
Brianna’s age: half of Aunt Anna’s 42 → 42 ÷ 2 = 21.  
Caitlin is 5 years younger than Brianna → 21 − 5 = 16.  
ANSWER 10: B

Problem 1:
Peter ate one full slice and shared another slice equally with his brother Paul.
- The pizza has 12 slices, so each slice is 1/12 of the pizza.
- Peter ate 1 full slice = 1/12 of the pizza.
- Peter shared 1 slice equally with Paul, so Peter ate half of that slice = (1/2) × (1/12) = 1/24 of the pizza.
- Total fraction Peter ate = 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8.

Ruling out other choices: 1/12 would be just one slice (ignoring the shared slice), 1/24 would be just the shared half-slice (ignoring the full slice), and 1/6 or 1/4 would overcount.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each.
- Total cost = 5 × $2.50 = $12.50.
- A 20%-off sale means he would save 20% of the total cost.
- Savings = 0.20 × $12.50 = $2.50.

Ruling out other choices: $1.00 would be 20% of just one folder's price ($5.00 × 0.20 = $1.00 is wrong since one folder is $2.50, not $5.00). $5.00 would be the cost of 2 folders. The correct savings is $2.50.

ANSWER 2: C

Problem 3:
A full circle has 360 degrees, which corresponds to 500 clerts. A right angle is 90 degrees, which is 90/360 = 1/4 of a full circle.
- Clerts in a right angle = (1/4) × 500 = 125 clerts.

Ruling out other choices: 90 would just be the degree measure, 250 would be half a circle, and 180 would be a straight angle in degrees. The proportional answer is 125.

ANSWER 3: C

Problem 4:
The choices appear to be fractions. For x = 7:
- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1.000
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Comparing all values: 0.75 < 0.857 < 1.000 < 1.167 < 1.333. The smallest is 6/(x+1) = 6/8 = 0.75.

Ruling out others: Choice A (6/7) is tempting but 6/8 is smaller because dividing 6 by a larger number gives a smaller result. Choices C, D, and E are all greater than or equal to 1.

ANSWER 4: B

Problem 5:
Numbers 1–64 are placed row by row on an 8×8 grid.
- Row 1: 1, 2, 3, 4, 5, 6, 7, 8
- Row 2: 9, 10, 11, 12, 13, 14, 15, 16
- ...
- Row 8: 57, 58, 59, 60, 61, 62, 63, 64

The four corners are:
- Top-left: 1
- Top-right: 8
- Bottom-left: 57 (first number of row 8: 7×8 + 1 = 57)
- Bottom-right: 64

Sum = 1 + 8 + 57 + 64 = 130.

Ruling out other choices: A common error is miscounting the bottom-left corner (e.g., using 56 instead of 57), which would give 129, not among the choices. The correct sum is 130.

ANSWER 5: A

Problem 6:
We need the number of digits in 4⁵ · 5¹⁰.
- Rewrite 4⁵ = (2²)⁵ = 2¹⁰.
- So 4⁵ · 5¹⁰ = 2¹⁰ · 5¹⁰ = (2 · 5)¹⁰ = 10¹⁰.
- 10¹⁰ = 10,000,000,000, which is a 1 followed by 10 zeros.
- This number has 11 digits.

Ruling out other choices: 10 might be tempting if one counts only the zeros, but the leading "1" makes it 11 digits total.

ANSWER 6: D

Problem 7:
First game: Annika + 3 friends = 4 players. Each gets 15 cards.
- Total cards in the deck = 4 × 15 = 60 cards.

Second game: 2 more friends join, so 4 + 2 = 6 players.
- Cards per player = 60 ÷ 6 = 10 cards.

Ruling out other choices: 12 would result from dividing by 5 players instead of 6. 8 would result from dividing by too many players. The correct answer with 6 players is 10.

ANSWER 7: C

Problem 8:
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)

Pair terms from the first group with terms from the second group (first with last, etc.):
- 1 + 49 = 50
- 11 + 39 = 50
- 21 + 29 = 50
- 31 + 19 = 50
- 41 + 9 = 50

Total = 5 × 50 = 250.

Alternatively, compute each sum: First group = 105, Second group = 145, Total = 250.

Ruling out other choices: 200 might result from a pairing error. 150 might come from computing only one group. The correct total is 250.

ANSWER 8: E

Problem 9:
Ahn picks a two-digit integer n, computes 200 − n, then doubles it: Result = 2(200 − n).
- To maximize the result, we must minimize n.
- The smallest two-digit integer is 10.
- Result = 2(200 − 10) = 2(190) = 380.

Ruling out other choices: 398 would come from using n = 1, but 1 is not a two-digit integer. 200 would come from n = 100, which is three digits. The largest valid result is 380.

ANSWER 9: D

Problem 10:
- Aunt Anna is 42 years old.
- Brianna is half as old as Aunt Anna: Brianna = 42 ÷ 2 = 21.
- Caitlin is 5 years younger than Brianna: Caitlin = 21 − 5 = 16.

Ruling out other choices: 15 might come from subtracting 6 instead of 5. 21 is Brianna's age, not Caitlin's. 37 is Aunt Anna's age minus 5. The correct age for Caitlin is 16.

ANSWER 10: B

 Problem 1:
What fraction of the whole pizza did Peter eat?  
The pizza has 12 slices. Peter ate one full slice, so that is \(1\) slice. He also shared another slice equally with his brother Paul, so Peter ate half of that second slice, which is \(\frac12\) slice. In total Peter ate  
\[
1+\frac12=\frac32\text{ slices.}
\]  
The fraction of the 12-slice pizza he ate is  
\[
\frac{3/2}{12}=\frac{3}{2\cdot12}=\frac{3}{24}=\frac18.
\]  
The tempting wrong answer \(\frac16\) would correspond to eating 2 slices, but Peter only ate \(1\frac12\).  

ANSWER 1: C

---

Problem 2:
How much could Karl have saved by waiting for the 20%-off sale?  
Karl bought 5 folders at \$2.50 each. The total cost is  
\[
5\times\$2.50=\$12.50.
\]  
A 20% discount on this total is  
\[
0.20\times\$12.50=\$2.50.
\]  
(Equivalently, each folder would be \$0.50 cheaper, and \(5\times\$0.50=\$2.50\).)  

ANSWER 2: C

---

Problem 3:
How many clerts are in a right angle if a full circle contains 500 clerts?  
A right angle is \(\frac14\) of a full circle. Therefore the number of clerts in a right angle is  
\[
\frac14\times500=125.
\]  

ANSWER 3: C

---

Problem 4:
For \(x=7\), which of the given expressions is the smallest?  
Reading the spaced notation as fractions:

- A. \(\dfrac{6}{x}=\dfrac{6}{7}\approx0.857\)
- B. \(\dfrac{6}{x+1}=\dfrac{6}{8}=0.75\)
- C. \(\dfrac{6}{x-1}=\dfrac{6}{6}=1\)
- D. \(\dfrac{x}{6}=\dfrac{7}{6}\approx1.167\)
- E. \(\dfrac{x+1}{6}=\dfrac{8}{6}\approx1.333\)

Comparing the values:  
\[
0.75<0.857<1<1.167<1.333.
\]  
Thus \(\dfrac{6}{x+1}\) is the smallest. A tempting mistake is to think \(\frac67\) is smaller because the denominator is smaller, but a smaller denominator actually makes a larger fraction here.  

ANSWER 4: B

---

Problem 5:
What is the sum of the numbers in the four corners of the \(8\times8\) checkerboard?  
The first row contains the numbers \(1,2,\dots,8\). Its corners are \(1\) (top-left) and \(8\) (top-right).  
Each row has 8 numbers. After 7 complete rows there are \(7\times8=56\) numbers, so the last row begins with \(57\) and ends with \(64\). Its corners are \(57\) (bottom-left) and \(64\) (bottom-right).  

The required sum is  
\[
1+8+57+64=130.
\]  
(Notice \(1+64=65\) and \(8+57=65\), so the total is \(2\times65=130\).)  

ANSWER 5: A

---

Problem 6:
How many digits are in the product \(4^5\cdot5^{10}\)?  
Rewrite \(4^5\) as \((2^2)^5=2^{10}\). Then  
\[
4^5\cdot5^{10}=2^{10}\cdot5^{10}=(2\cdot5)^{10}=10^{10}.
\]  
\(10^{10}\) is written as a \(1\) followed by ten zeros: \(10\,000\,000\,000\).  
Counting the digits gives \(1+10=11\).  

ANSWER 6: D

---

Problem 7:
How many cards will each player receive when 2 more friends join?  
Annika and 3 friends make \(4\) players. With 15 cards each, the total number of cards in the deck is  
\[
4\times15=60.
\]  
If 2 more friends join, there are \(4+2=6\) players. Dividing the same 60 cards evenly gives  
\[
60\div6=10\text{ cards per player.}
\]  

ANSWER 7: C

---

Problem 8:
Compute \((1+11+21+31+41)+(9+19+29+39+49)\).  
Pair the terms conveniently:  

\[
\begin{aligned}
&(1+49)+(11+39)+(21+29)+(31+19)+(41+9)\\
&=50+50+50+50+50\\
&=5\times50=250.
\end{aligned}
\]  

(As a check, the first group sums to \(105\) and the second to \(145\); \(105+145=250\).)  

ANSWER 8: E

---

Problem 9:
What is the largest result Ahn can obtain?  
Let the two-digit integer chosen be \(n\), where \(10\le n\le 99\). The operation produces  
\[
2(200-n)=400-2n.
\]  
To maximize this expression, we minimize \(n\). The smallest two-digit integer is \(10\). Substituting gives  
\[
(200-10)\times2=190\times2=380.
\]  
Choice \(398\) is a trap that comes from using the illegitimate “two-digit” number \(1\).  

ANSWER 9: D

---

Problem 10:
How old is Caitlin?  
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2}=21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21-5=16.
\]  

ANSWER 10: B

Problem 1:
Peter ate one full slice and half of another slice (shared equally with Paul). So he ate \(1 + \frac{1}{2} = \frac{3}{2}\) slices. The pizza has 12 slices, so the fraction is \(\frac{3/2}{12} = \frac{3}{24} = \frac{1}{8}\).  
ANSWER 1: C

Problem 2:
Karl bought 5 folders at $2.50 each, total cost = \(5 \times 2.50 = \$12.50\). A 20% discount would save \(0.20 \times 12.50 = \$2.50\).  
ANSWER 2: C

Problem 3:
A full circle = 500 clerts. A right angle is \(\frac{1}{4}\) of a full circle, so clerts = \(500 \times \frac{1}{4} = 125\).  
ANSWER 3: C

Problem 4:
For \(x = 7\), evaluate each expression:  
A. \(6x = 6 \times 7 = 42\)  
B. \(6x + 1 = 42 + 1 = 43\)  
C. \(6x - 1 = 42 - 1 = 41\)  
D. \(\frac{x}{6} = \frac{7}{6} \approx 1.1667\)  
E. \(\frac{x+1}{6} = \frac{8}{6} \approx 1.3333\)  
The smallest is \(\frac{7}{6}\), which is choice D.  
ANSWER 4: D

Problem 5:
The checkerboard is filled row by row with numbers 1 to 64.  
First row: squares 1–8, so top‑left corner = 1.  
Top‑right corner = 8.  
Last row: numbers 57–64, so bottom‑left corner = 57, bottom‑right corner = 64.  
Sum = \(1 + 8 + 57 + 64 = 130\).  
ANSWER 5: A

Problem 6:
Compute \(4^5 \cdot 5^{10}\). Rewrite as \(4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}\).  
\(10^{10}\) is a 1 followed by 10 zeros, so it has 11 digits.  
ANSWER 6: D

Problem 7:
Initially 4 players (Annika + 3 friends) each get 15 cards, so total cards = \(4 \times 15 = 60\).  
With 2 more friends, total players = 6. Cards per player = \(60 \div 6 = 10\).  
ANSWER 7: C

Problem 8:
Group the terms: \((1+9) + (11+19) + (21+29) + (31+39) + (41+49)\).  
Each pair sums to 10, 30, 50, 70, 90 respectively. Add them: \(10+30+50+70+90 = 250\).  
Alternatively, note each column sums to 50, and there are 5 columns, so total = 250.  
ANSWER 8: E

Problem 9:
Ahn chooses a two‑digit integer \(n\) (10 ≤ n ≤ 99). He computes \(200 - n\) and doubles it: \(2(200 - n) = 400 - 2n\). To maximize this, minimize \(n\). The smallest two‑digit integer is 10, giving \(400 - 20 = 380\).  
Check: if \(n=10\), result = 380. Larger \(n\) gives smaller result.  
ANSWER 9: D

Problem 10:
Aunt Anna = 42. Brianna is half as old as Aunt Anna: \(42 \div 2 = 21\). Caitlin is 5 years younger than Brianna: \(21 - 5 = 16\).  
ANSWER 10: B

Problem 1:
Peter ate one whole slice, which is \( \frac{1}{12} \) of the pizza.  
He also shared another slice equally with his brother, so he ate half of that slice:  
\[
\frac{1}{2}\times\frac{1}{12}=\frac{1}{24}.
\]  
Total fraction Peter ate:
\[
\frac{1}{12}+\frac{1}{24}=\frac{2}{24}+\frac{1}{24}=\frac{3}{24}=\frac{1}{8}.
\]  
Thus the answer is \(\frac{1}{8}\), choice C.

ANSWER 1: C

Problem 2:
Karl bought 5 folders at \$2.50 each, so his cost was  
\[
5 \times \$2.50 = \$12.50.
\]  
A 20 % discount on the total would be  
\[
0.20 \times \$12.50 = \$2.50.
\]  
Waiting for the sale would have saved him \$2.50, which corresponds to choice C.

ANSWER 2: C

Problem 3:
A full circle contains 500 clerts and corresponds to 360 degrees.  
A right angle is \(90^\circ\).  
Number of clerts in a right angle:
\[
90^\circ \times \frac{500\ \text{clerts}}{360^\circ}= \frac{90 \times 500}{360}=125.
\]  
Thus the answer is 125 clerts, choice C.

ANSWER 3: C

Problem 4:
For \(x = 7\) evaluate each expression:

- \(6x = 6\cdot7 = 42\)  
- \(6x+1 = 42+1 = 43\)  
- \(6x-1 = 42-1 = 41\)  
- \(\frac{x}{6} = \frac{7}{6} \approx 1.167\)  
- \(\frac{x+1}{6} = \frac{8}{6} \approx 1.333\)

The smallest value is \(\frac{x}{6}\), which is choice D.

ANSWER 4: D

Problem 5:
The numbers are placed row‑wise on an 8 × 8 board.  
- Top‑left corner: row 1, column 1 → number 1.  
- Top‑right corner: row 1, column 8 → number 8.  
- Bottom‑left corner: row 8, column 1 → number 57 (since row 8 contains 57–64).  
- Bottom‑right corner: row 8, column 8 → number 64.

Sum of the four corners:
\[
1 + 8 + 57 + 64 = 130.
\]  
Thus the answer is 130, choice A.

ANSWER 5: A

Problem 6:
\[
4^5 \cdot 5^{10} = (2^2)^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2\cdot5)^{10} = 10^{10}.
\]  
\(10^{10}\) is a 1 followed by 10 zeros, which contains \(11\) digits.  
Hence the answer is 11 digits, choice D.

ANSWER 6: D

Problem 7:
With Annika and 3 friends (4 players) each receiving 15 cards, the total number of cards is  
\[
4 \times 15 = 60.
\]  
If 2 more friends join, there are 6 players sharing the same 60 cards, so each gets  
\[
\frac{60}{6} = 10
\]  
cards.  
Thus the answer is 10, choice C.

ANSWER 7: C

Problem 8:
First sum: \(1 + 11 + 21 + 31 + 41\).  
This is an arithmetic series with 5 terms, first = 1, last = 41:  
\[
\frac{5}{2}(1+41) = \frac{5}{2}\times42 = 105.
\]  

Second sum: \(9 + 19 + 29 + 39 + 49\).  
Similarly, first = 9, last = 49:  
\[
\frac{5}{2}(9+49) = \frac{5}{2}\times58 = 145.
\]  

Total: \(105 + 145 = 250\).  
Thus the answer is 250, choice E.

ANSWER 8: E

Problem 9:
Let the chosen two‑digit integer be \(n\) (with \(10 \le n \le 99\)).  
Ahn computes \(2(200 - n) = 400 - 2n\).  
To maximize this expression, minimize \(n\). The smallest two‑digit integer is 10, giving  
\[
400 - 2\cdot10 = 380.
\]  
Thus the largest possible result is 380, choice D.

ANSWER 9: D

Problem 10:
Aunt Anna is 42. Brianna is half as old:  
\[
\frac{42}{2} = 21.
\]  
Caitlin is 5 years younger than Brianna:  
\[
21 - 5 = 16.
\]  
Hence Caitlin is 16 years old, choice B.

ANSWER 10: B

Problem 1:  
Peter’s family ordered a 12-slice pizza. Peter ate one full slice and shared another slice equally with his brother Paul. Each slice is $ \frac{1}{12} $ of the pizza. Eating one slice gives Peter $ \frac{1}{12} $. Sharing another slice means Peter eats half of $ \frac{1}{12} $, or $ \frac{1}{24} $. Total eaten: $ \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8} $.  
Tempting wrong choices: 1/24 (only the shared part), 1/12 (only one slice).  
ANSWER 1: C  

Problem 2:  
Karl bought 5 folders at $2.50 each, so total cost = $5 \times 2.50 = 12.50$. The next day, 20% off applies. Savings per folder = $2.50 \times 0.20 = 0.50$. Total savings for 5 folders = $5 \times 0.50 = 2.50$.  
Tempting wrong choices: $1.00 (underestimates 20% of total), $2.00 (incorrect per-folder calculation).  
ANSWER 2: C  

Problem 3:  
A full circle is 500 clerts. A right angle is $ \frac{1}{4} $ of a full circle. Clerts in a right angle = $ \frac{500}{4} = 125 $.  
Tempting wrong choices: 90 (Earth angles), 100 (misdivides 500 by 5).  
ANSWER 3: C  

Problem 4:  
For $ x = 7 $:  
- A: $ \frac{6}{7} \approx 0.857 $  
- B: $ \frac{6}{7+1} = \frac{6}{8} = 0.75 $  
- C: $ \frac{6}{7-1} = \frac{6}{6} = 1 $  
- D: $ \frac{7}{6} \approx 1.167 $  
- E: $ \frac{7+1}{6} = \frac{8}{6} \approx 1.333 $  
Smallest is $ \frac{6}{8} = 0.75 $ (B).  
Tempting wrong choices: D or E (larger numerators/denominators).  
ANSWER 4: A (since 6/7 ≈0.857 is larger than 6/(7+1)=0.75, the smallest is B? Wait no, rechecking: A is 6/7 ≈0.857, B is 6/8=0.75, which is smaller. So B is smallest.  
Correction: The smallest is B. Original reasoning had a typo.  
**Revised**: For x=7, A=6/7≈0.857, B=6/8=0.75, C=6/6=1, D=7/6≈1.167, E=8/6≈1.333. Smallest is B (0.75).  
Tempting wrong choices: A (close to B but larger).  
ANSWER 4: B  

Problem 5:  
Checkerboard numbers 1 to 64. Corners:  
- Top-left: 1  
- Top-right: 8  
- Bottom-left: 57 (first of 8th row: 57–64)  
- Bottom-right: 64  
Sum: $1 + 8 + 57 + 64 = 130$.  
Tempting wrong choices: 131–134 (misadd or misidentify corners).  
ANSWER 5: A  

Problem 6:  
Compute $4^5 \times 5^{10}$:  
- $4^5 = (2^2)^5 = 2^{10} = 1024$  
- $5^{10} = 9765625$  
- Product: $1024 \times 9765625 = 9990000000...$ Wait, better: $4^5 \times 5^{10} = 2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$.  
$10^{10}$ has 11 digits (1 followed by 10 zeros).  
Tempting wrong choices: 8–10 (underestimates exponent).  
ANSWER 6: A (No, 10^10 is 1 followed by 10 zeros, so 11 digits).  
**Correction**: $10^{10}$ is 11 digits.  
ANSWER 6: A? No, choices are A.8 B.9 C.10 D.11 E.12. So D.  
**Final**: $10^{10}$ has 11 digits.  
ANSWER 6: A (No, D).  
**Final Answer**: D  

Problem 7:  
Total cards: $4 \times 15 = 60$. If 2 more join, total players = 6. Cards per player: $60 \div 6 = 10$.  
Tempting wrong choices: 8–12 (misdivides total cards).  
ANSWER 7: C  

Problem 8:  
Compute:  
$(1 + 11 + 21 + 31 + 41) = 105$  
$(9 + 19 + 29 + 39 + 49) = 145$  
Total: $105 + 145 = 250$.  
Tempting wrong choices: 150–249 (partial sums or misadds).  
ANSWER 8: E  

Problem 9:  
Let the two-digit integer be $n$ ($10 \leq n \leq 99$). Ahn computes $2(200 - n)$. To maximize, minimize $n$. Smallest $n = 10$. Result: $2(200 - 10) = 380$.  
Tempting wrong choices: 200–220 (misinterprets operation).  
ANSWER 9: D  

Problem 10:  
Aunt Anna = 42. Brianna = $42 \div 2 = 21$. Caitlin = $21 - 5 = 16$.  
Tempting wrong choices: 15 (subtracts 6), 17 (adds instead), 21 (Brianna’s age).  
ANSWER 10: B  


**Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: A (No, D) **Final: D**  
**Correction for 6**: $10^{10}$ has 11 digits.  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B  

**Corrected Final Answers**  
ANSWER 1: C  
ANSWER 2: C  
ANSWER 3: C  
ANSWER 4: B  
ANSWER 5: A  
ANSWER 6: D  
ANSWER 7: C  
ANSWER 8: E  
ANSWER 9: D  
ANSWER 10: B

### Problem 1:
What is the fraction of the whole pizza Peter ate, given the 12-slice pizza: he ate 1 full slice and half of another shared slice?
First, calculate the number of slices Peter ate: 1 full slice + 0.5 shared slices = 1.5 total slices. The pizza has 12 slices total, so the fraction is $\frac{1.5}{12} = \frac{3}{24} = \frac{1}{8}$.
Trap wrong answers: A (1/24, far too small), B (1/12, only counts the full slice), D (1/6, counts 2 full slices).
ANSWER 1: C

---

### Problem 2:
How much would Karl save by buying 5 $2.50 folders during a 20% off sale?
First calculate Karl's original total spend: $5 \times 2.50 = \$12.50$. 20% off means he saves 20% of the total, so $0.2 \times 12.50 = \$2.50$. Alternatively, he saves $0.50 per folder (20% of $2.50), times 5 folders = $2.50 total savings.
Trap wrong answers: A ($1, miscalculates savings per folder), E ($5, calculates 40% savings instead of 20%).
ANSWER 2: C

---

### Problem 3:
A full circle is 500 clerts. How many clerts are in a right angle?
A right angle is 1/4 of a full circle (the same as 90 degrees of a 360-degree standard circle). For a 500-clert full circle, a right angle is $\frac{500}{4} = 125$ clerts.
Trap wrong answers: A (90, uses degrees instead of clerts), E (250, half a circle, a straight angle not a right angle).
ANSWER 3: C

---

### Problem 4:
For x=7, which value is the smallest? (Options are formatted as fractions, with spacing errors corrected: A. $\frac{6}{x}$, B. $\frac{6}{x+1}$, C. $\frac{6}{x-1}$, D. $\frac{x}{6}$, E. $\frac{x+1}{6}$)
Calculate each value:
A: $\frac{6}{7} \approx 0.857$
B: $\frac{6}{8} = 0.75$
C: $\frac{6}{6} = 1$
D: $\frac{7}{6} \approx 1.167$
E: $\frac{8}{6} \approx 1.333$
The smallest value is 0.75 from option B.
Trap wrong answers: A (0.857, larger than B's value), C (1, much larger).
ANSWER 4: B

---

### Problem 5:
Numbers 1-64 are written row-wise on an 8x8 grid (first row 1-8, second 9-16, etc.). What is the sum of the four corner numbers?
First identify the four corners:
1. Top-left (row 1, column 1): 1
2. Top-right (row 1, column 8): 8
3. Bottom-left (row 8, column 1): $7 \times 8 +1 = 57$ (7 full rows of 8 numbers before the 8th row)
4. Bottom-right (row 8, column 8): 64
Sum: $1 + 8 + 57 + 64 = 130$.
Trap wrong answers: B (131, miscalculates the bottom-left corner as 58).
ANSWER 5: A

---

### Problem 6:
How many digits are in the product $4^5 \times 5^{10}$?
Rewrite the product to simplify: $4^5 = (2^2)^5 = 2^{10}$, so the product becomes $2^{10} \times 5^{10} = (2 \times 5)^{10} = 10^{10}$. $10^{10}$ is 1 followed by 10 zeros, which is 11 total digits.
Trap wrong answers: C (10, confuses the exponent 10 with the number of digits).
ANSWER 6: D

---

### Problem 7:
Annika plus 3 friends (4 total players) get 15 cards each in a card game. If 2 more players join the next game, how many cards does each player get, if the same total number of cards is split evenly?
First calculate total number of cards: $4 \times 15 = 60$ total cards. The new number of players is $4 + 2 = 6$. Each player gets $\frac{60}{6} = 10$ cards.
Trap wrong answers: B (9, miscounts the original number of players as 3 instead of 4).
ANSWER 7: C

---

### Problem 8:
Calculate the sum: $(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49)$
Pair terms to simplify: $(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 + 50 + 50 + 50 + 50 = 250$.
Trap wrong answers: C (200, miscalculates the number of 50-pairs as 4 instead of 5).
ANSWER 8: E

---

### Problem 9:
Ahn picks a two-digit integer N, calculates $2 \times (200 - N)$. What is the largest possible value of this number?
To maximize the result, we need to minimize N (since it is subtracted from 200). The smallest two-digit integer is 10 (two-digit integers range from 10 to 99, 0-9 are one-digit). Substitute N=10: $2 \times (200 -10) = 2 \times 190 = 380$.
Trap wrong answers: E (398, uses the one-digit integer N=1 which is invalid), A (200, uses the three-digit integer N=100 which is invalid).
ANSWER 9: D

---

### Problem 10:
Aunt Anna is 42. Brianna is half Anna's age, and Caitlin is 5 years younger than Brianna. How old is Caitlin?
First calculate Brianna's age: $42 / 2 = 21$ years old. Caitlin's age is $21 -5 = 16$ years old.
Trap wrong answers: D (21, Brianna's age not Caitlin's), E (37, Anna's age minus 5, incorrectly assigning the 5-year gap to Anna).
ANSWER 10: B

Problem 1:
We need to find the fraction of the 12-slice pizza Peter ate. Peter ate 1 full slice, which is $\frac{1}{12}$ of the pizza. He shared a second slice equally with his brother, so he ate $\frac{1}{2}$ of that slice, which is $\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}$ of the pizza. Adding these together:
$$\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$$
ANSWER 1: C
Problem 2:
First calculate the total cost of Karl's original purchase: 5 folders × $2.50 = $12.50. The 20% discount applies to the original price, so the savings are 20% of $12.50:
$$0.20 \times 12.50 = 2.50$$
Alternatively, the discount per folder is $2.50 × 0.20 = $0.50, so total savings for 5 folders is 5 × $0.50 = $2.50.
ANSWER 2: C
Problem 3:
A right angle is $\frac{1}{4}$ of a full circle. Since a full circle is 500 clerts, a right angle is:
$$\frac{500}{4} = 125$$
ANSWER 3: C
Problem 4:
Substitute $x=7$ into each expression:
- A. $6x = 6×7 = 42$
- B. $6x +1 = 42 +1 =43$
- C. $6x -1 =42 -1=41$
- D. $x^6 =7^6=117649$ (interpreting "x 6" as $x^6$, consistent with AMC 8 formatting)
- E. $(x+1)^6=8^6=262144$ (interpreting "x +1 6" as $(x+1)^6$)
The smallest value is 41, from option C.
ANSWER 4: C
Problem 5:
The four corners of the 8×8 checkerboard are:
- Top-left: 1 (first number)
- Top-right: 8 (last number of first row)
- Bottom-left: 57 (first number of 8th row: $8×7 +1=57$)
- Bottom-right: 64 (last number of 8th row)
Sum the corners:
$$1 + 8 + 57 + 64 = (1+64) + (8+57) = 65 + 65 = 130$$
ANSWER 5: A
Problem 6:
Rewrite the product using exponent rules to simplify:
$$4^5 \times 5^{10} = (2^2)^5 \times 5^{10} = 2^{10} \times 5^{10} = (2×5)^{10} = 10^{10}$$
$10^{10}$ is 1 followed by 10 zeros, which has $10 +1 =11$ digits.
ANSWER 6: D
Problem 7:
First find the total number of cards: 4 players × 15 cards each = 60 cards. When 2 more friends join, there are $4+2=6$ players total. The number of cards per player is:
$$\frac{60}{6}=10$$
ANSWER 7: C
Problem 8:
Add the two groups of numbers. We can pair terms to simplify:
$$(1+49) + (11+39) + (21+29) + (31+19) + (41+9) = 50 +50 +50 +50 +50 = 5×50=250$$
Alternatively, sum each arithmetic sequence separately:
- First group: $\frac{5}{2}×(1+41) = 105$
- Second group: $\frac{5}{2}×(9+49)=145$
- Total: $105+145=250$
ANSWER 8: E
Problem 9:
Let the two-digit integer be $n$. The expression is $2×(200 -n) = 400 - 2n$. To maximize this value, we minimize $n$ (since we subtract $2n$). The smallest positive two-digit integer is 10. Substitute $n=10$:
$$2×(200 -10) = 2×190 = 380$$
(If negative two-digit integers were allowed, the result would be larger, but no such option exists, so we assume positive two-digit integers as standard for AMC 8.)
ANSWER 9: D
Problem 10:
Brianna is half as old as Aunt Anna: $\frac{42}{2}=21$ years old. Caitlin is 5 years younger than Brianna: $21 -5=16$ years old.
ANSWER 10: B