🔬 Model Lab

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel

Problem 1:
We are asked for the value of a 3×3×3 cube given a 2×2×2 cube is worth $200.

Value is proportional to amount of silver, i.e. to weight, i.e. to volume.

Volume of 2-inch cube = $2^3 = 8$ cubic inches.
Volume of 3-inch cube = $3^3 = 27$ cubic inches.

Ratio = $27/8$.

So value of larger = $200 \times 27/8 = 25 \times 27$.

$25 \times 27 = 25 \times 20 + 25 \times 7 = 500+175 = 675$.

Choice A $300$ would be linear scaling by side length $3/2$. Wrong because value scales as cube. Choice C $450$ etc. are wrong.

ANSWER 1: E

Problem 2:
We need which card must be turned to disprove: "If a vowel is on one side, then an even number is on the other side."

This is an If P then Q statement with P = vowel, Q = even.
To show it false we need P true and Q false, i.e. a card with vowel on one side and odd on the other.

Cards showing: P, Q, 3, 4, 6. Each has letter on one side, number on the other.

- P showing: letter P is a consonant, so P is false. The rule says nothing about consonants. Turning it cannot disprove.
- Q showing: letter Q is a consonant, so same. No use.
- 4 showing: number 4 is even, so Q is true. Even with a vowel behind it would confirm, with a consonant it is irrelevant. Cannot disprove.
- 6 showing: same as 4, even, cannot disprove.
- 3 showing: number 3 is odd, so Q is false. If the other side is a vowel, then we have vowel + odd, disproving the rule. So this is the card to check.

Thus Mary turned over 3.

ANSWER 2: A

Problem 3:
Let total money be $S$, constant through giving.

Toy starts with $36$. After Amy doubles Jan and Toy, Toy has $72$.
After Jan doubles Amy and Toy, Toy has $144$.

Call that $T_2 = 144$.

In the last step, Toy gives Amy and Jan enough to double them, and is left with $36$.

So amount Toy gave away = $144-36 = 108$.
That amount equals $A_2 + J_2$, the amounts Amy and Jan had before doubling.

After doubling, they have $2A_2 + 2J_2 = 2 \times 108 = 216$.

Total $S$ = what Toy has left + what Amy and Jan now have
$= 36 + 216 = 252$.

Thus total is $252$.

ANSWER 3: D

Problem 4:
We have 7 numbers in order. First four are positions 1,2,3,4. Last four are 4,5,6,7. The 4th number is counted in both.

Sum of first four = average $\times$ count = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum with double count = $20+32 = 52$.

True sum of all seven = $7 \times 6\frac47$.
$6\frac47 = 46/7$. Times 7 = $46$.

Difference $52-46 = 6$ is the extra counting of the common number, i.e. the 4th number itself.

So common number = $6$.

ANSWER 4: B

Problem 5:
Buy: 4 for $5, so cost per disk = $5/4$.
Sell: 3 for $5, so revenue per disk = $5/3$.

Profit per disk = $5/3 - 5/4 = 20/12 - 15/12 = 5/12$ dollars.

Another way: for 12 disks, cost = $3 \times 5 = \$15$, revenue = $4 \times 5 = \$20$, profit = $\$5$.

Need profit $\$100$. Number of 12-disk lots = $100/5 = 20$.
Disks = $20 \times 12 = 240$.

Check: $240 \times 5/12 = 100$.

ANSWER 5: D

Problem 6:
Sum $1+2+\dots+9 = 45$. For 3 equal-sum groups, each sum = $45/3 = 15$.

List triples summing to 15:
1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.

The number 9 must be with 1+5 or 2+4.

Case 1: group $1,5,9$. Remaining $\{2,3,4,6,7,8\}$.
The only triples from the list using only these are $2+6+7=15$ and $3+4+8=15$, which exactly partition the remainder. So one partition:
$\{1,5,9\},\{2,6,7\},\{3,4,8\}$.

Case 2: group $2,4,9$. Remaining $\{1,3,5,6,7,8\}$.
The only triples are $1+6+8=15$ and $3+5+7=15$, which partition the remainder. So second partition:
$\{2,4,9\},\{1,6,8\},\{3,5,7\}$.

No other possibility. Up to order of groups, there are 2 ways.

ANSWER 6: C

Problem 7:
Start with 128.

Sell 25% to Jill: $25\%$ of $128 = 32$. Remaining = $128-32 = 96$.

Sell 25% of remaining to June: $25\%$ of $96 = 24$. Remaining = $96-24 = 72$.

Give 1 to teacher: $72-1 = 71$.

So Jack has 71.

ANSWER 7: D

Problem 8:
Distribute 24 identical apples to Alice, Becky, Chris, each $\ge 2$.

Let $a,b,c$ be their numbers, $a+b+c=24$, $a,b,c\ge 2$.

Put $a'=a-2$, etc. Then $a',b',c'\ge 0$ and
$a'+b'+c' = 24-6 = 18$.

Number of nonnegative solutions to $x+y+z=18$ is stars-and-bars:
$\binom{18+3-1}{3-1} = \binom{20}{2} = 20\cdot19/2 = 190$.

People are distinct, so each triple counts.

ANSWER 8: C

Problem 9:
Let A,B,C,D be "Alan gets A", etc.
Statements: $A \Rightarrow B$, $B \Rightarrow C$, $C \Rightarrow D$, all true. Exactly two of A,B,C,D true.

If A true, then B true, then C true, then D true: 4 trues, impossible.
So A false.

If B true, then C true, then D true: at least 3 trues, impossible with exactly 2.
So B false.

Thus A,B false. The two As must include D, because if D false, the two trues would be among A,B,C and would force D true. So D true.

The other true cannot be A or B, so must be C. Check: C true, D true gives: $A\Rightarrow B$ vacuously true since A false; $B\Rightarrow C$ vacuously true since B false; $C\Rightarrow D$ true since both true. Works.

So Carlos and Diana.

ANSWER 9: C

Problem 10:
Color follows place. Moving changes color.

3 greens leave shade for sun: -3 green, +3 yellow.
5 yellows leave sun for shade: +5 green, -5 yellow.

Net: green $+2$, yellow $-2$. Total number of frogs unchanged.

Let initially $G_0,Y_0$ with $G_0/Y_0 = 3/1$, so $G_0=3Y_0$.
Now $G_1 = G_0+2$, $Y_1 = Y_0-2$, with $G_1/Y_1 = 4/1$.

So $G_0+2 = 4(Y_0-2)$.
$3Y_0+2 = 4Y_0-8$.
$Y_0 = 10$.

Then $G_0 = 30$. Now $G_1 = 32$, $Y_1 = 8$.
Difference now = $32-8 = 24$.

ANSWER 10: E

Problem 1:
**Restatement of the problem:** We are given that a 2-inch cube of silver weighs 3 pounds and costs $200. We need to find the value of a 3-inch cube of silver.

**Mathematical setup and computation:**
The value of a solid cube of silver is directly proportional to its mass, which is proportional to its volume.
1. The volume of the 2-inch cube is:
   $$\text{Volume}_1 = 2 \times 2 \times 2 = 8\text{ cubic inches.}$$
2. The volume of the 3-inch cube is:
   $$\text{Volume}_2 = 3 \times 3 \times 3 = 27\text{ cubic inches.}$$
3. Since an 8-cubic-inch cube is worth $\$200$, the value per cubic inch is:
   $$\frac{\$200}{8} = \$25\text{ per cubic inch.}$$
4. Therefore, the value of the 27-cubic-inch cube is:
   $$27 \times \$25 = \$675.$$

**Ruling out wrong choices:**
A linear scaling factor of $\frac{3}{2}$ would give $\$200 \times \frac{3}{2} = \$300$ (Choice A), and an area scaling factor of $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ would give $\$450$ (Choice C), both of which incorrectly ignore the 3-dimensional nature of volume.

ANSWER 1: E

---

Problem 2:
**Restatement of the problem:** Five cards show $P, Q, 3, 4, 6$. Each card has a letter on one side and a whole number on the other side. Jane claims: "If a vowel is on one side of any card, then an even number is on the other side." Mary disproved this by turning over just one card. We must determine which card Mary turned over.

**Mathematical setup and computation:**
A conditional statement of the form "If $A$, then $B$" is only falsified when the hypothesis $A$ is true and the conclusion $B$ is false:
- $A$: "The card has a vowel on one side."
- $\text{not } B$: "The card has an odd number on the other side."

Thus, to disprove Jane's statement, Mary needs a card that has both a **vowel** and an **odd number**.
- Looking at the cards showing letters: $P$ and $Q$ are consonants. Even if either had an odd number on the other side, it would not violate the rule because the hypothesis (vowel) is not met.
- Looking at the cards showing numbers: 
  - $4$ and $6$ are even numbers. Whether the other side is a vowel or consonant, the conditional statement is not violated.
  - $3$ is an odd number. If Mary turns over the card showing $3$ and finds a vowel, Jane's rule is disproved. 

Therefore, Mary must have turned over the card with $3$.

ANSWER 2: A

---

Problem 3:
**Restatement of the problem:** Three friends, Amy, Jan, and Toy, redistribute their money in three rounds:
1. Amy doubles Jan's and Toy's amounts.
2. Jan doubles Amy's and Toy's amounts.
3. Toy doubles Amy's and Jan's amounts.
Toy starts with $\$36$ and ends with $\$36$. We want to find the total amount of money among all three friends.

**Mathematical setup and computation:**
The total amount of money $S$ is conserved throughout the entire process.
- Let Toy's initial money be $T_0 = 36$.
- **Round 1 (Amy gives):** Toy's amount doubles, so:
  $$T_1 = 2 \times T_0 = 2 \times 36 = 72.$$
- **Round 2 (Jan gives):** Toy's amount doubles again, so:
  $$T_2 = 2 \times T_1 = 2 \times 72 = 144.$$
- **Round 3 (Toy gives):** Toy gives Amy and Jan enough to double each of their amounts.
  Let $A_2$ and $J_2$ be Amy's and Jan's amounts after Round 2.
  To double their money, Toy must give Amy $A_2$ and Jan $J_2$.
  Thus, Toy gives away $A_2 + J_2$.
  Toy's remaining money is:
  $$T_3 = T_2 - (A_2 + J_2).$$
  Since the total sum of money is $S = A_2 + J_2 + T_2$, we have:
  $$A_2 + J_2 = S - T_2 = S - 144.$$
  Substituting this into the expression for $T_3$:
  $$T_3 = 144 - (S - 144) = 288 - S.$$
  We are given that $T_3 = 36$:
  $$288 - S = 36 \implies S = 288 - 36 = 252.$$

**Ruling out wrong choices:**
The total must be $\$252$. Answers like $\$108$ ($3 \times 36$) assume equal starting amounts, which is not stated and leads to a contradiction.

ANSWER 3: D

---

Problem 4:
**Restatement of the problem:** A list of seven numbers has the properties:
- The average of the first four numbers is $5$.
- The average of the last four numbers is $8$.
- The average of all seven numbers is $6\frac{4}{7}$.
We need to find the number that is common to both sets of four numbers (the 4th number).

**Mathematical setup and computation:**
Let the seven numbers in order be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$.
- The sum of the first four numbers is:
  $$x_1 + x_2 + x_3 + x_4 = 4 \times 5 = 20.$$
- The sum of the last four numbers is:
  $$x_4 + x_5 + x_6 + x_7 = 4 \times 8 = 32.$$
- The sum of all seven numbers is:
  $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46.$$

Adding the sum of the first four and the sum of the last four counts the common number $x_4$ twice:
$$(x_1 + x_2 + x_3 + x_4) + (x_4 + x_5 + x_6 + x_7) = (\text{Sum of all 7}) + x_4$$
$$20 + 32 = 46 + x_4$$
$$52 = 46 + x_4 \implies x_4 = 6.$$

ANSWER 4: B

---

Problem 5:
**Restatement of the problem:** Maria buys computer disks at $4$ for $\$5$ and sells them at $3$ for $\$5$. We need to find how many disks she must sell to make a profit of $\$100$.

**Mathematical setup and computation:**
- Maria's cost per disk is:
  $$\text{Cost} = \frac{\$5}{4} = \$1.25.$$
- Maria's selling price per disk is:
  $$\text{Revenue} = \frac{\$5}{3}.$$
- The profit per disk sold is:
  $$\text{Profit per disk} = \frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}\text{ dollars}.$$
- To earn a total profit of $\$100$, the number of disks $N$ she must sell satisfies:
  $$N \times \frac{5}{12} = 100 \implies N = 100 \times \frac{12}{5} = 20 \times 12 = 240.$$

**Ruling out wrong choices:**
A common mistake is thinking the profit is $\$1$ per 12 disks, which gives $1200$ (Choice E), but the profit for 12 disks is actually $12 \times \frac{5}{12} = \$5$. Dividing $\$100$ by $\$5$ gives 20 sets of 12 disks, or $20 \times 12 = 240$ disks.

ANSWER 5: D

---

Problem 6:
**Restatement of the problem:** The numbers $1, 2, \dots, 9$ are written on cards. Alina divides the 9 cards into 3 groups of 3 card

# Problem 1
**Asked:** Find the value of a 3-inch silver cube, given a 2-inch cube is worth $200.

Silver is priced by **amount of material**, i.e., by volume (or equivalently weight), not by edge length.

- 2-inch cube: volume = 2 × 2 × 2 = 8 cubic inches → worth $200
- 3-inch cube: volume = 3 × 3 × 3 = 27 cubic inches

Value scales with volume:
$$200 \times \frac{27}{8} = 200 \times 3.375 = \$675$$

**Ruling out traps:**
- $300 comes from scaling by edge length (200 × 3/2) — wrong, since a cube 1.5× longer holds much more than 1.5× the material.
- $450 comes from scaling by surface area (200 × 9/4) — wrong, we pay for volume, not surface.

ANSWER 1: E

# Problem 2
**Asked:** Which single card, when flipped, can prove Jane wrong?

Jane's claim: **"vowel on one side ⇒ even number on the other side."** This is one-directional. To refute it, we must find a card with a **vowel paired with an odd number**. Check each card:

- **Flip 4 or 6:** reveals a letter. If it's a vowel, that vowel has an *even* number on the back — claim holds. If it's a consonant, the claim doesn't apply. **Can never refute Jane.**
- **Flip P or Q:** reveals a number. But P and Q are **consonants**, and the claim only restricts cards with vowels. No number behind them can violate it. **Can never refute Jane.**
- **Flip 3:** reveals a letter. If that letter is a **vowel**, then a vowel has the odd number 3 on its other side — **Jane is wrong!**

Since Mary *did* prove Jane wrong with one card, it must be the card that is capable of showing a violation: the **3**.

ANSWER 2: A

# Problem 3
**Asked:** Find the total money, given Toy starts and ends with $36.

Key facts: money only moves among the three friends, so the total $T$ never changes. Track **Toy**:

- Start: Toy has $36.
- After Amy doubles everyone: Toy has $72.
- After Jan doubles everyone: Toy has $144.
- **Toy's turn:** he gives away exactly (Amy's current amount + Jan's current amount). At that moment Amy + Jan together hold $T - 144$ (since all three hold $T$ and Toy holds $144$). So Toy ends with:
$$144 - (T - 144) = 288 - T$$

Toy ends with $36:
$$288 - T = 36 \implies T = 252$$

**Check with a valid example:** Amy $150, Jan $66, Toy $36 (total $252).
- Amy gives $102: Amy 48, Jan 132, Toy 72.
- Jan gives $120: Amy 96, Jan 12, Toy 144.
- Toy gives $108: Amy 192, Jan 24, Toy **36** ✓.

**Trap:** $216 (choice C) is Amy + Jan's combined amount — forgetting to add Toy's $36.

ANSWER 3: D

# Problem 4
**Asked:** Find the number counted in both "the first four" and "the last four."

Convert the averages into sums:
- First four numbers: sum = 4 × 5 = 20
- Last four numbers: sum = 4 × 8 = 32
- All seven numbers: sum = 7 × (6 4/7) = 7 × 46/7 = 46

The middle (4th) number is the **only overlap** — it is included in both the "first four" and the "last four," so it gets counted **twice** in 20 + 32. Therefore:
$$20 + 32 - x = 46 \implies 52 - x = 46 \implies x = 6$$

**Trap:** 6 4/7 (choice C) is just the overall average; the overlap number is exactly 6.

ANSWER 4: B

# Problem 5
**Asked:** How many disks must she sell to profit $100?

Compute profit **per disk**:
- Cost per disk: $5 ÷ 4 = $5/4
- Selling price per disk: $5 ÷ 3 = $5/3
- Profit per disk: $\frac{5}{3} - \frac{5}{4} = \frac{20 - 15}{12} = \frac{5}{12}$ dollar

Number of disks needed:
$$n \times \frac{5}{12} = 100 \implies n = \frac{1200}{5} = 240$$

**Check with dozens:** Buy 12 disks for $15; sell 12 for $20 → profit $5 per dozen. For $100 profit: 20 dozen = **240 disks** ✓.

**Trap:** 1200 comes from dividing by the profit per *some wrong unit*; 200 or 100 ignore that she must sell in groups matching the price structure — the per-disk method is exact.

ANSWER 5: D

# Problem 6
**Asked:** Number of ways to split {1, …, 9} into 3 groups of 3 with equal sums.

Total = 1+2+…+9 = 45, so each group must sum to **15**.

List all 3-number subsets of {1,…,9} summing to 15 (these are exactly the 8 "lines" of the 3×3 magic square):
$$\{1,5,9\},\ \{1,6,8\},\ \{2,4,9\},\ \{2,5,8\},\ \{2,6,7\},\ \{3,4,8\},\ \{3,5,7\},\ \{4,5,6\}$$

Every partition must use exactly one triple containing **1**, which is either {1,5,9} or {1,6,8}:

- **Using {1,5,9}:** remaining numbers {2,3,4,6,7,8}. The only listed triples avoiding 1, 5, 9 are {2,6,7} and {3,4,8}, which exactly cover the remainder. → 1 way.
- **Using {1,6,8}:** remaining numbers {2,3,4,5,7,9}. The only listed triples avoiding 1, 6, 8 are {2,4,9} and {3,5,7}, which exactly cover the remainder. → 1 way.

Total: **2 ways** (the groups are unlabeled, so order doesn't matter):
- {1,5,9}, {2,6,7}, {3,4,8}
- {1,6,8}, {2,4,9}, {3,5,7}

ANSWER 6: C

# Problem 7
**Asked:** How many apples remain after the sales and gift?

Work step by step:
- Start: 128 apples.
- Sold 25% to Jill: 25% of 128 = 32 sold → 128 − 32 = **96 remain**.
- Sold 25% of *those remaining* to June: 25% of 96 = 24 sold → 96 − 24 = **72 remain**.
- Gave 1 to teacher: 72 − 1 = **71**.

**Trap:** 72 (not a choice here, but the common slip) forgets the teacher's apple; the answer is 71.

ANSWER 7: D

# Problem 8
**Asked:** Ways for Alice, Becky, Chris to split 24 apples, each getting at least 2.

Let Alice get $a$, Becky $b$, Chris $c$ with
$$a + b + c = 24,\qquad a, b, c \ge 2$$

Substitute $a = a' + 2$, $b = b' + 2$, $c = c' + 2$ (so $a', b', c' \ge 0$):
$$a' + b' + c' = 18$$

Number of nonnegative integer solutions (stars and bars):
$$\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \frac{20 \cdot 19}{2} = 190$$

**Trap:** 380 = 2 × 190 would double-count (e.g., treating orderings as distinct); 210 comes from miscounting the shift by 2.

ANSWER 8: C

# Problem 9
**Asked:** Which two students got A's, given all three implications are true?

The chain: **Alan ⇒ Beth ⇒ Carlos ⇒ Diana.** Test each option against "exactly two A's":

- **(A) Alan, Beth:** Beth's A forces Carlos's A → 3 A's ✗
- **(B) Beth, Carlos:** Carlos's A forces Diana's A → 3 A's ✗
- **(C) Carlos, Diana:** Alan has no A, so "If Alan gets an A…" i

Problem 1:
The value of a silver cube is proportional to its volume, since the density and price per weight are the same.

A 2-inch cube has volume  
\[
2^3=8
\]
cubic inches.

A 3-inch cube has volume  
\[
3^3=27
\]
cubic inches.

So the 3-inch cube has  
\[
\frac{27}{8}
\]
times as much silver as the 2-inch cube. Since the 2-inch cube is worth \$200, the 3-inch cube is worth  
\[
200\cdot \frac{27}{8}=25\cdot 27=675.
\]

The tempting answer \$300 would come from scaling by side length, but volume scales by the cube of the side length.

ANSWER 1: E

Problem 2:
Jane’s statement is:

\[
\text{If a card has a vowel on one side, then it has an even number on the other side.}
\]

To show the statement false, Mary needs to find a card with a vowel on one side and an odd number on the other.

The visible cards are:
\[
P,\ Q,\ 3,\ 4,\ 6.
\]

- Cards showing consonants \(P\) and \(Q\) cannot violate the statement, because the visible side is not a vowel.
- Cards showing even numbers \(4\) and \(6\) cannot violate the statement, because if the other side is a vowel, the number side is even.
- The card showing \(3\) is the only odd number. If its other side is a vowel, then a vowel is paired with an odd number, which disproves Jane’s claim.

So Mary turned over the card showing \(3\).

ANSWER 2: A

Problem 3:
Let Amy’s initial amount be \(a\), Jan’s initial amount be \(b\), and Toy’s initial amount be \(36\).

After Amy gives enough to double Jan and Toy:
- Jan doubles from \(b\) to \(2b\).
- Toy doubles from \(36\) to \(72\).
- Amy loses \(b+36\), so Amy has
\[
a-b-36.
\]

So after the first step:
\[
(a-b-36,\ 2b,\ 72).
\]

Next, Jan gives enough to double Amy and Toy:
- Amy doubles from \(a-b-36\) to
\[
2a-2b-72.
\]
- Toy doubles from \(72\) to \(144\).
- Jan loses \((a-b-36)+72\), so Jan has
\[
2b-(a-b-36)-72=3b-a-36.
\]

So after the second step:
\[
(2a-2b-72,\ 3b-a-36,\ 144).
\]

Finally, Toy gives enough to double Amy and Jan. Toy gives away
\[
(2a-2b-72)+(3b-a-36)=a+b-108.
\]

Thus Toy’s final amount is
\[
144-(a+b-108)=252-a-b.
\]

We are told Toy ends with \$36, so
\[
252-a-b=36.
\]
Therefore,
\[
a+b=216.
\]

The total amount the three friends have is
\[
a+b+36=216+36=252.
\]

ANSWER 3: D

Problem 4:
The average of the first four numbers is 5, so their sum is
\[
4\cdot 5=20.
\]

The average of the last four numbers is 8, so their sum is
\[
4\cdot 8=32.
\]

The average of all seven numbers is
\[
6\frac47=\frac{46}{7},
\]
so the sum of all seven numbers is
\[
7\cdot \frac{46}{7}=46.
\]

The first four numbers and the last four numbers overlap in exactly one number: the fourth number. When we add the two sums \(20\) and \(32\), every number is counted once except the common number, which is counted twice.

So if the common number is \(x\), then
\[
20+32=46+x.
\]
Thus,
\[
52=46+x,
\]
so
\[
x=6.
\]

ANSWER 4: B

Problem 5:
Maria buys 4 disks for \$5, so she buys in batches of 4.

She sells 3 disks for \$5, so she sells in batches of 3.

Use a common batch size of 12 disks:
- Buying 12 disks costs
\[
3\cdot 5=15.
\]
- Selling 12 disks earns
\[
4\cdot 5=20.
\]
- Profit on 12 disks is
\[
20-15=5.
\]

She needs a profit of \$100, so she needs
\[
\frac{100}{5}=20
\]
batches of 12 disks.

Therefore, the number of disks she must sell is
\[
20\cdot 12=240.
\]

ANSWER 5: D

Problem 6:
The numbers on the cards are
\[
1,2,3,4,5,6,7,8,9.
\]

Their total sum is
\[
1+2+\cdots+9=45.
\]

Since the cards are divided into 3 groups of 3 with equal sums, each group must sum to
\[
45\div 3=15.
\]

Now count the ways to split the numbers into three groups of three, each summing to 15.

The number 9 must be in one of the groups. The other two numbers in its group must sum to 6. The only possibilities are
\[
9,1,5
\]
or
\[
9,2,4.
\]

Case 1: The group containing 9 is \(\{1,5,9\}\).

The remaining numbers are
\[
2,3,4,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{2,6,7\}\quad\text{and}\quad \{3,4,8\}.
\]

So one valid division is
\[
\{1,5,9\},\ \{2,6,7\},\ \{3,4,8\}.
\]

Case 2: The group containing 9 is \(\{2,4,9\}\).

The remaining numbers are
\[
1,3,5,6,7,8.
\]

These must form two groups summing to 15. The only way is
\[
\{1,6,8\}\quad\text{and}\quad \{3,5,7\}.
\]

So a second valid division is
\[
\{2,4,9\},\ \{1,6,8\},\ \{3,5,7\}.
\]

There are exactly 2 ways.

ANSWER 6: C

Problem 7:
Jack starts with 128 apples.

He sells 25% to Jill, so he keeps 75%:
\[
128\cdot 0.75=96.
\]

He then sells 25% of the remaining apples to June, so again he keeps 75%:
\[
96\cdot 0.75=72.
\]

Then he gives 1 apple to his teacher:
\[
72-1=71.
\]

ANSWER 7: D

Problem 8:
Let the number of apples Alice, Becky, and Chris receive be \(a\), \(b\), and \(c\), respectively.

We need
\[
a+b+c=24
\]
with
\[
a,b,c\ge 2.
\]

Subtract 2 from each person’s share:
\[
a'=a-2,\qquad b'=b-2,\qquad c'=c-2.
\]

Then
\[
a'+b'+c'=(24)-6=18,
\]
where \(a',b',c'\) are nonnegative integers.

The number of nonnegative integer solutions to
\[
a'+b'+c'=18
\]
is given by stars and bars:
\[
\binom{18+3-1}{3-1}=\binom{20}{2}.
\]

Compute:
\[
\binom{20}{2}=\frac{20\cdot 19}{2}=190.
\]

ANSWER 8: C

Problem 9:
The statements are:

\[
\text{Alan A} \implies \text{Beth A}
\]
\[
\text{Beth A} \implies \text{Carlos A}
\]
\[
\text{Carlos A} \implies \text{Diana A}
\]

So if Alan gets an A, then Beth, Carlos, and Diana must also get A’s. That would make 4 students with A’s, which is impossible because only two students received A’s.

Therefore, Alan did not get an A.

If Beth gets an A, then Carlos and Diana must also get A’s. That would make 3 students with A’s, still impossible.

Therefore, Beth did not get an A.

The only possible way to have exactly two A’s is for Carlos and Diana to get A’s, with Alan and Beth not getting A’s.

This works because:
- Alan’s statement is vacuously true since Alan did not get an A.
- Beth’s statement is vacuously true since Beth did not get an A.
- Carlos’s statement is true because Carlos got an A and Dia

Problem 1:
A 2-inch cube has volume 2³ = 8 cubic inches; a 3-inch cube has volume 3³ = 27 cubic inches.
Value is proportional to the amount of silver (volume/weight), so:

Value = $200 × (27/8) = $200 × 3.375 = $675.

(The tempting error is scaling by the side ratio 3/2, giving $300, or by the area ratio 9/4, giving $450 — but silver is sold by weight, which follows volume.)

ANSWER 1: E

Problem 2:
Jane's claim: "vowel on one side ⟹ even number on other side."
To disprove it, Mary must find a card with a vowel paired with an odd number.
- Cards P and Q show consonants — flipping them shows a number, but the letter side isn't a vowel, so the rule can't be broken.
- Cards 4 and 6 show even numbers — whatever letter is behind them, the rule "vowel ⟹ even" is satisfied.
- Card 3 shows an odd number. If a vowel is behind it, the rule fails. This is the only card that could reveal a counterexample.

ANSWER 2: A

Problem 3:
Let the total be T (constant, since money is only passed around). Toy starts with $36.
- Amy doubles Jan's and Toy's amounts: Toy now has 2 × 36 = $72.
- Jan doubles Amy's and Toy's amounts: Toy now has 2 × 72 = $144.
- Toy doubles Amy's and Jan's amounts: if Amy + Jan = S just before this step, Toy pays out S, ending with 144 − S = 36, so S = 108.

After the last step, Amy + Jan = 2S = 216, and Toy has 36.
Total T = 216 + 36 = $252.

ANSWER 3: D

Problem 4:
Sum of first four = 4 × 5 = 20. Sum of last four = 4 × 8 = 32.
Sum of all seven = 7 × 6 4/7 = 7 × 46/7 = 46.
Adding the two groups of four counts the middle (4th) number twice:
20 + 32 = 46 + (common number) ⟹ 52 − 46 = 6.

ANSWER 4: B

Problem 5:
Cost per disk = 5/4 = $1.25. Selling price per disk = 5/3 ≈ $1.667.
Profit per disk = 5/3 − 5/4 = 20/12 − 15/12 = 5/12 dollars.
Disks needed: 100 ÷ (5/12) = 100 × 12/5 = 240.

(Choice 120 comes from mistakenly using a profit of $5/6 per disk; 1200 from ignoring the fraction.)

ANSWER 5: D

Problem 6:
Total = 1 + 2 + ⋯ + 9 = 45, so each group of 3 must sum to 45/3 = 15.
All triples from 1–9 summing to 15: {1,5,9}, {1,6,8}, {2,4,9}, {2,5,8}, {2,6,7}, {3,4,8}, {3,5,7}, {4,5,6}.
Consider the group containing 1: it is {1,5,9} or {1,6,8}.
- {1,5,9}: remaining {2,3,4,6,7,8} must split into two triples of sum 15: only {2,6,7} and {3,4,8}. ✔ (1 way)
- {1,6,8}: remaining {2,3,4,5,7,9}: only {2,4,9} and {3,5,7}. ✔ (1 way)

So there are exactly 2 ways.

ANSWER 6: C

Problem 7:
Sold to Jill: 25% of 128 = 32, leaving 128 − 32 = 96.
Sold to June: 25% of 96 = 24, leaving 96 − 24 = 72.
Gave 1 to the teacher: 72 − 1 = 71.

(63 would come from subtracting 25% of 128 twice; 65 from a similar slip.)

ANSWER 7: D

Problem 8:
Let a, b, c be the numbers Alice, Becky, and Chris get, with a + b + c = 24 and each ≥ 2.
Give each person 2 apples first: 24 − 6 = 18 apples remain to distribute freely.
Let a′ = a − 2, etc., with a′ + b′ + c′ = 18, each ≥ 0.
Number of nonnegative solutions = C(18 + 2, 2) = C(20, 2) = 190.

ANSWER 8: C

Problem 9:
The statements are conditionals; from "Alan ⟹ Beth ⟹ Carlos ⟹ Diana," anyone getting an A forces everyone later in the chain to get an A.
- If Alan got an A, then Beth, Carlos, Diana all did (4 A's) — too many.
- If Beth got an A, then Carlos and Diana did (at least 3) — too many.
- If Carlos got an A, then Diana did: exactly Carlos and Diana can be the only two, and all three statements stay true (Alan's and Beth's are true since they didn't get A's).

So Carlos and Diana received the A's.

ANSWER 9: C

Problem 10:
Let the initial numbers be green = 3y, yellow = y.
Three greens move to sun (turn yellow) and five yellows move to shade (turn green):
New green = 3y − 3 + 5 = 3y + 2; new yellow = y + 3 − 5 = y − 2.
Set the new ratio: (3y + 2)/(y − 2) = 4 ⟹ 3y + 2 = 4y − 8 ⟹ y = 10.
Now green = 32, yellow = 8 (total still 40 ✓).
Difference = 32 − 8 = 24.

ANSWER 10: E

Problem 1:

A \(2\times2\times2\) cube has volume

\[
2^3=8\text{ cubic inches}.
\]

A \(3\times3\times3\) cube has volume

\[
3^3=27\text{ cubic inches}.
\]

Since both are silver, value is proportional to volume. Thus the three-inch cube is worth

\[
200\cdot\frac{27}{8}=25\cdot27=675.
\]

A tempting error is to compare side lengths, using \(\frac32\), but the value depends on volume, so the factor is \(\left(\frac32\right)^3\).

\[
\boxed{\$675}
\]

ANSWER 1: E

Problem 2:

Jane’s statement is:

> If a card has a vowel on one side, then it has an even number on the other.

To prove Jane wrong, Mary must find a card with:

- a vowel on one side, and
- an odd number on the other side.

The visible odd number is \(3\). If the other side of the \(3\) card is a vowel, Jane’s statement is false.

Turning over \(4\) or \(6\) cannot disprove the rule, because even numbers are allowed on the other side of vowels. Turning over \(P\) or \(Q\) also cannot disprove the rule, since they are consonants and the rule says nothing about consonants.

Therefore Mary turned over the \(3\) card.

ANSWER 2: A

Problem 3:

We are told Toy begins with \(\$36\).

After Amy’s turn, Amy gives Toy enough to double Toy’s money:

\[
36\to72.
\]

Then Jan gives Toy enough to double Toy’s money again:

\[
72\to144.
\]

So, just before Toy gives money to Amy and Jan, Toy has \(\$144\).

Let Amy have \(x\) dollars after Amy’s first turn. Jan then doubles Amy’s amount, so Amy has \(2x\) dollars before Toy’s turn.

Let Jan have \(y\) dollars after Jan gives money to Amy and Toy.

On Toy’s turn:

- Toy gives Amy \(2x\) dollars to double Amy’s money.
- Toy gives Jan \(y\) dollars to double Jan’s money.

Toy ends with \(\$36\), so Toy gave away

\[
144-36=108.
\]

Thus,

\[
2x+y=108.
\]

Just before Toy’s final transaction, the total money is

\[
2x+y+144.
\]

Using \(2x+y=108\),

\[
108+144=252.
\]

Money is only redistributed, not created or destroyed, so the total amount is \(\$252\).

A tempting mistake is to add only Toy’s beginning and ending amounts, but the question asks for all three friends’ money.

ANSWER 3: D

Problem 4:

The average of the first four numbers is \(5\), so their sum is

\[
4\cdot5=20.
\]

The average of the last four numbers is \(8\), so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47\), so the total of all seven numbers is

\[
7\left(6\frac47\right)
=7\left(\frac{46}{7}\right)
=46.
\]

When we add the first-four sum and last-four sum, the number common to both sets is counted twice:

\[
20+32=52.
\]

Since the true total is \(46\), the repeated number is

\[
52-46=6.
\]

A tempting error is to say the common number is \(20+32\), but that total includes all seven numbers as well as one extra copy of the shared number.

ANSWER 4: B

Problem 5:

Maria buys disks at \(4\) for \(\$5\), so each disk costs

\[
\frac54=\$1.25.
\]

She sells disks at \(3\) for \(\$5\), so each disk sells for

\[
\frac53=\$1.\overline{6}.
\]

Her profit per disk is

\[
\frac53-\frac54
=\frac{20}{12}-\frac{15}{12}
=\frac{5}{12}.
\]

To make \(\$100\) profit, she must sell

\[
100\div\frac{5}{12}
=100\cdot\frac{12}{5}
=240
\]

disks.

A tempting wrong answer is \(120\), obtained by confusing total sales revenue with profit. Profit is only the amount left after paying for the disks.

ANSWER 5: D

Problem 6:

The sum of the numbers from \(1\) through \(9\) is

\[
1+2+\cdots+9=45.
\]

Since there are three groups with equal sums, each group must have sum

\[
45\div3=15.
\]

The possible triples of distinct numbers that sum to \(15\) are:

\[
(1,5,9),\ (1,6,8),\ (2,4,9),\ (2,5,8),
\]
\[
(2,6,7),\ (3,4,8),\ (3,5,7),\ (4,5,6).
\]

Since \(9\) must be in some group, there are only two possibilities for its group:

\[
(1,5,9)
\quad\text{or}\quad
(2,4,9).
\]

If the group is \((1,5,9)\), the remaining cards form

\[
(2,6,7)\quad\text{and}\quad(3,4,8).
\]

This gives one division:

\[
\{1,5,9\},\{2,6,7\},\{3,4,8\}.
\]

If the group is \((2,4,9)\), the remaining cards form

\[
(1,6,8)\quad\text{and}\quad(3,5,7).
\]

This gives a second division:

\[
\{2,4,9\},\{1,6,8\},\{3,5,7\}.
\]

Thus there are \(2\) ways. Rearranging the groups does not create a new division.

ANSWER 6: C

Problem 7:

Jack starts with \(128\) apples.

He sells \(25\%\), so he keeps \(75\%\):

\[
128\cdot\frac34=96.
\]

Then he sells \(25\%\) of the \(96\) remaining apples, so he again keeps \(75\%\):

\[
96\cdot\frac34=72.
\]

Finally, he gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract \(25\%\) of the original \(128\) twice. The second \(25\%\) is taken from the remaining \(96\), not from the original amount.

ANSWER 7: D

Problem 8:

Let the numbers of apples Alice, Becky, and Chris receive be \(a,b,c\), respectively.

They satisfy

\[
a+b+c=24,
\]

and each person must receive at least \(2\) apples.

Give each person \(2\) apples first. This uses

\[
2+2+2=6
\]

apples, leaving

\[
24-6=18
\]

apples to distribute freely.

The number of nonnegative solutions to

\[
x+y+z=18
\]

is

\[
\binom{18+3-1}{3-1}
=
\binom{20}{2}
=
\frac{20\cdot19}{2}
=
190.
\]

The people are named, so giving Alice \(5\), Becky \(7\), and Chris \(12\) is different from giving Alice \(7\), Becky \(5\), and Chris \(12\).

ANSWER 8: C

Problem 9:

The statements are:

\[
\text{Alan gets A} \implies \text{Beth gets A},
\]

\[
\text{Beth gets A} \implies \text{Carlos gets A},
\]

\[
\text{Carlos gets A} \implies \text{Diana gets A}.
\]

Thus, if Alan got an A, then all four would get A’s. That is impossible because exactly two students got A’s.

If Beth got an A, then Beth, Carlos, and Diana would all get A’s. That is also impossible.

Therefore Alan and Beth did not get A’s.

If Carlos got an A, then Diana must also get an A. This gives exactly two A’s:

\[
\text{Carlos and Diana}.
\]

Choices involving Alan or Beth would force too many people to receive A’s.

ANSWER 9: C

Problem 10:

Initially the ratio of green frogs to yel