🔬 Model Lab

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make 

Problem 1:
We are asked for the average of the first and last numbers after rearranging $-2,4,6,9,12$ with conditions.

Sorted: smallest is $-2$, median of 5 numbers is $6$, largest is $12$.

Let positions be 1 to 5.
Largest $12$ is not first, but in first three places: position in $\{1,2,3\}$, not $1$, so in $\{2,3\}$.
Smallest $-2$ is not last, but in last three places: position in $\{3,4,5\}$, not $5$, so in $\{3,4\}$.
Median $6$ is not first or last: position in $\{2,3,4\}$.

Hence:
$12$ cannot be in position 1 or 5, because it must be in $2$ or $3$.
$-2$ cannot be in position 1 or 5, because it must be in $3$ or $4$.
$6$ cannot be in position 1 or 5, by rule.

So position 1 cannot be $-2,6,12$. It must be $4$ or $9$.
Similarly position 5 cannot be $-2,6,12$. It must be $4$ or $9$.

Thus the first and last numbers are $4$ and $9$ in some order. Their average is:
$$(4+9)/2 = 13/2 = 6.5$$

This is feasible, e.g. $4,12,6,-2,9$ satisfies all rules. The other choices do not give $13/2$.
ANSWER 1: C

Problem 2:
We need the capacity $C$ in gallons of a tank that goes from $1/3$ full to $1/2$ full when $4$ gallons are added.

Let $C$ = capacity.
Amount before $= C/3$.
Amount after $= C/3 + 4$.
This equals $C/2$.

So:
$$C/3 + 4 = C/2$$
$$4 = C/2 - C/3 = C/6$$
$$C = 24$$

Check: $1/3$ of $24$ is $8$; $8+4=12$ which is $1/2$ of $24$. Choices like $12$ or $48$ do not satisfy this.
ANSWER 2: D

Problem 3:
Ralph bought $12$ pairs for $\$24$, with prices $\$1,\$3,\$4$, at least one of each. Find number of $\$1$ pairs.

Let $x$ = $\$1$ pairs, $y$ = $\$3$ pairs, $z$ = $\$4$ pairs.
Then:
$$x+y+z = 12$$
$$x+3y+4z = 24$$
$x\ge 1, y\ge 1, z\ge 1$, integers.

Subtract first from second:
$$(x+3y+4z)-(x+y+z) = 24-12$$
$$2y+3z = 12$$

So $2y = 12-3z$. Hence $3z$ must be even, so $z$ is even.
Also $y\ge 1$ so $3z \le 10$, and $z\ge 1$. So $z=2$ is the only even possibility.

Then $2y+6=12$, so $2y=6$, $y=3$.
Then $x = 12-y-z = 12-3-2 = 7$.

Check: $7\cdot1+3\cdot3+2\cdot4 = 7+9+8=24$. Other values fail the equations.
ANSWER 3: D

Problem 4:
Gas expands $4$ cubic cm for every $3^\circ$ rise. Volume is $24$ at $32^\circ$. Find volume at $20^\circ$.

Temperature change from $32^\circ$ to $20^\circ$:
$$32-20 = 12^\circ \text{ drop}$$

Number of $3^\circ$ intervals:
$$12/3 = 4$$

Each interval is $4$ cc. Drop means shrink:
$$4 \times 4 = 16 \text{ cc decrease}$$

So volume at $20^\circ$:
$$24-16 = 8$$

Choices like $40$ would be for a rise; $16$ ignores starting volume.
ANSWER 4: A

Problem 5:
Numbers are $99$-digit $3030\dots0303$ and $5050\dots0505$. Their product has thousands digit $A$ and units digit $B$. Find $A+B$.

Units digit: units digits are $3$ and $5$. $3\times5=15$, so units digit $B=5$.

For thousands digit, only last $4$ digits matter, since higher digits contribute multiples of $10000$.

The first number alternates $3,0,\dots,3$. From the right: units $3$, tens $0$, hundreds $3$, thousands $0$. So last four digits are $0303$, i.e. $303$ mod $10000$.
Similarly second number ends $0505$, i.e. $505$ mod $10000$.

Then product mod $10000$:
$$303 \times 505 = 303\times(500+5)=151500+1515=153015$$
$$153015 \bmod 10000 = 3015$$

So last four digits of product are $3015$. Thousands digit $A=3$.

Sum:
$$A+B = 3+5 = 8$$

Not $5,6,$ etc.
ANSWER 5: D

Problem 6:
Find the largest possible quotient using two numbers from $\{-24,-3,-2,1,2,8\}$.

A quotient $a/b$ is positive if $a,b$ have same sign, negative otherwise. The largest will be positive.

Positive/positive with $1,2,8$: largest is $8/1=8$.

Negative/negative with $-24,-3,-2$: to maximize, use largest magnitude numerator over smallest magnitude denominator: $(-24)/(-2)=12$. Other combos: $(-24)/(-3)=8$, $(-3)/(-2)=1.5$, etc., all $\le 12$.

Mixed signs give negative quotients, less than positive ones.

So the maximum is $12$. $24$ is not attainable; $-24$ is the minimum.
ANSWER 6: D

Problem 7:
May salary $\$2000$. June $20\%$ raise, July $20\%$ cut. Find final salary.

June after $20\%$ raise:
$$2000 \times 1.20 = 2400$$

July after $20\%$ cut of the new amount:
$$2400 \times 0.80 = 1920$$

It is not $\$2000$, because the $20\%$ cut is taken on the larger $\$2400$ base, so it removes more than was added.
ANSWER 7: A

Problem 8:
$3$-inch cube made of $27$ $1$-inch cubes, $19$ white, $8$ black at corners. What fraction of large surface is white?

Large surface area in unit squares: $6$ faces $\times 3\times3 =54$ unit squares.

A corner small cube exposes $3$ faces. With $8$ black corner cubes:
$$\text{black exposed} = 8\times3 =24$$

Total exposed $54$, so white exposed:
$$54-24 =30$$

Fraction white:
$$30/54 = 5/9$$

It is not $19/27$, which is the volume fraction of white cubes; surface exposure differs.
ANSWER 8: D

Problem 9:
Before district: won $45\%$. District: $6$ wins, $2$ losses. End: won $1/2$ of all games. Find total games.

Let $T$ = total games. District games $=6+2=8$. Games before $=T-8$.

Wins before $=0.45(T-8)$. Total wins $=0.45(T-8)+6 = 0.5T$.

Solve:
$$0.45T -3.6 +6 = 0.5T$$
$$2.4 = 0.05T$$
$$T = 2.4/0.05 = 48$$

Check: before $40$ games, $18$ wins $=45\%$; total $24$ wins out of $48$ $=50\%$.
ANSWER 9: A

Problem 10:
Colfax $270$ students, boys:girls $5:4$. Winthrop $180$ students, boys:girls $4:5$. All attend dance. What fraction are girls?

Colfax parts $5+4=9$. $270/9=30$ per part.
Boys $5\times30=150$, girls $4\times30=120$.

Winthrop parts $4+5=9$. $180/9=20$ per part.
Boys $4\times20=80$, girls $5\times20=100$.

Totals: students $270+180=450$. Girls $120+100=220$.

Fraction girls:
$$220/450 = 22/45$$

Not $1/2$; Colfax has more students and fewer girls proportionally.
ANSWER 10: C

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

---

### Problem 1:
**Restatement:**
We are given the five numbers $-2, 4, 6, 9, 12$. They are arranged in positions 1 to 5 according to three rules:
1. The largest number ($12$) is not first, but is in one of the first three positions (positions 2 or 3).
2. The smallest number ($-2$) is not last, but is in one of the last three positions (positions 3 or 4).
3. The median number ($6$) is not first or last (positions 2, 3, or 4).

We need to find the average of the first and last numbers (positions 1 and 5).

**Step-by-step working:**
- The five positions are 1, 2, 3, 4, 5.
- The number $12$ can only be in position 2 or 3. Therefore, $12$ cannot be in position 1 or 5.
- The number $-2$ can only be in position 3 or 4. Therefore, $-2$ cannot be in position 1 or 5.
- The number $6$ can only be in position 2, 3, or 4. Therefore, $6$ cannot be in position 1 or 5.

This completely excludes $-2$, $6$, and $12$ from being either the first or the last number. The only remaining numbers are $4$ and $9$. Thus, one of the positions 1 and 5 must be $4$, and the other must be $9$.

The average of the first and last numbers is:
$$\frac{4 + 9}{2} = \frac{13}{2} = 6.5$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Adding 4 gallons of liquid to a tank that is $\frac{1}{3}$ full makes it $\frac{1}{2}$ full. We need to find the total capacity of the tank in gallons.

**Step-by-step working:**
Let $C$ be the total capacity of the tank in gallons.
- The initial amount of liquid is $\frac{1}{3}C$.
- Adding 4 gallons gives $\frac{1}{3}C + 4$.
- The tank is now $\frac{1}{2}$ full, so:
$$\frac{1}{3}C + 4 = \frac{1}{2}C$$

Subtract $\frac{1}{3}C$ from both sides:
$$4 = \left(\frac{1}{2} - \frac{1}{3}\right)C$$
$$4 = \left(\frac{3}{6} - \frac{2}{6}\right)C = \frac{1}{6}C$$

Multiply both sides by 6:
$$C = 4 \times 6 = 24\text{ gallons}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
Ralph bought 12 pairs of socks for a total of $\$24$. The socks cost $\$1$, $\$3$, and $\$4$ per pair. He bought at least one pair of each type. We need to find how many pairs of $\$1$ socks he bought.

**Step-by-step working:**
Let $x$ be the number of $\$1$ pairs, $y$ be the number of $\$3$ pairs, and $z$ be the number of $\$4$ pairs.
We have:
1) $x + y + z = 12$
2) $x + 3y + 4z = 24$
with $x, y, z \ge 1$ being integers.

Subtract equation (1) from equation (2):
$$(x + 3y + 4z) - (x + y + z) = 24 - 12$$
$$2y + 3z = 12$$

Since $y$ and $z$ are positive integers:
- If $z = 1$: $2y = 12 - 3(1) = 9$, which has no integer solution for $y$.
- If $z = 2$: $2y = 12 - 3(2) = 6 \implies y = 3$.
- If $z \ge 3$: $2y = 12 - 3(3) = 3$ (not an even integer), and for $z \ge 4$, $2y \le 0$, which contradicts $y \ge 1$.

Thus, the only solution is $z = 2$ and $y = 3$.
Substitute back to find $x$:
$$x = 12 - y - z = 12 - 3 - 2 = 7$$

Ralph bought 7 pairs of $\$1$ socks.

ANSWER 3: D

---

### Problem 4:
**Restatement:**
The gas expands by $4\text{ cm}^3$ for every $3^\circ$ rise in temperature. The volume is $24\text{ cm}^3$ at $32^\circ$. We need to find the volume at $20^\circ$.

**Step-by-step working:**
- The difference in temperature from $20^\circ$ to $32^\circ$ is:
$$32^\circ - 20^\circ = 12^\circ$$
- The number of $3^\circ$ increments in $12^\circ$ is:
$$\frac{12^\circ}{3^\circ} = 4\text{ increments}$$
- In 4 increments of $3^\circ$, the volume increases by:
$$4 \times 4\text{ cm}^3 = 16\text{ cm}^3$$
- Therefore, the volume at $20^\circ$ was the volume at $32^\circ$ minus the expansion:
$$24\text{ cm}^3 - 16\text{ cm}^3 = 8\text{ cm}^3$$

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Let $X = 303,030,\dots,030,303$ and $Y = 505,050,\dots,050,505$ be two 99-digit numbers. We need to find the sum of the thousands digit $A$ and the units digit $B$ of the product $X \times Y$.

**Step-by-step working:**
The thousands and units digits of a product depend only on the last four digits of the factors (modulo $10,000$).
- Looking at the end of $X$, the last digits are $\dots,030,303$. Thus, the last four digits of $X$ are $0303$, so:
$$X \equiv 303 \pmod{10000}$$
- Looking at the end of $Y$, the last digits are $\dots,050,505$. Thus, the last four digits of $Y$ are $0505$, so:
$$Y \equiv 505 \pmod{10000}$$

Now, compute the product modulo $10,000$:
$$303 \times 505 = 303 \times (500 + 5) = 151500 + 1515 = 153015$$
Taking this modulo $10,000$ gives the last four digits:
$$153015 \equiv 3015 \pmod{10000}$$

Reading the digits of $3015$:
- Thousands digit $A = 3$
- Hundreds digit $= 0$
- Tens digit $= 1$
- Units digit $B = 5$

The sum is:
$$A + B = 3 + 5 = 8$$

ANSWER 5: D

---

### Problem 6:
**Restatement:**
Find the largest quotient $\frac{a}{b}$ that can be formed using two numbers $a, b$ from the set $\{-24, -3, -2, 1, 2, 8\}$.

**Step-by-step working:**
To maximize the quotient, we want the result to be positive and as large as possible. A positive quotient comes from either two positive numbers or two negative numbers:
1. Positive divided by positive:
   - $\frac{8}{1} = 8$
   - $\frac{8}{2} = 4$
   - $\frac{2}{1} = 2$
2. Negative divided by negative:
   - $\frac{-24}{-2} = 12$
   - $\frac{-24}{-3} = 8$
   - $\frac{-3}{-2} = 1.5$

Comparing all positive results, the maximum is:
$$\frac{-24}{-2} = 12$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Ana's salary was $\$2000$ in May. She got a $20\%$ raise in June, followed by a $20\%$ pay cut in July. We want to find her salary after these two changes.

**Step-by-step working:**
- A $20\%$ raise means multiplying by $(1 + 0.20) = 1.20$:
$$\text{June salary} = 2000 \times 1.20 = 2400\text{ dollars}$$
- A $20\%$ cut on the new salary means multiplying by $(1 - 0.20) = 0.80$:
$$\text{July salary} = 2400 \times 0.80 = 1920\text{ dollars}$$

Alternatively, the combined multiplier is $1.20 \times 0.80 = 0.96$, and $2000 \times 0.96 = 1920$ dollars.

ANSWER 7: A

---

### Problem 8:
**Restatement:**
A $3 \times 3 \times 3$ cube consists

# Problem 1:

The numbers are −2, 4, 6, 9, 12. In order: largest = **12**, smallest = **−2**, median = **6**.

Label the five positions 1, 2, 3, 4, 5. Translate the rules:
- 12 is in one of the first three places but not first → 12 is in position **2 or 3**
- −2 is in one of the last three places but not last → −2 is in position **3 or 4**
- 6 is not first or last → 6 is in position **2, 3, or 4**

**Case 1: 12 is in position 2.** Then −2 must be in position 3 or 4, and 6 must be in position 3 or 4 (position 2 is taken). So −2 and 6 fill positions 3 and 4, leaving 4 and 9 for positions 1 and 5.

**Case 2: 12 is in position 3.** Then −2 must be in position 4 (position 3 is taken), and 6 must be in position 2 (positions 3 and 4 are taken). Again, 4 and 9 are left for positions 1 and 5.

In every possible arrangement, the first and last numbers are **4 and 9** (in some order). Their average is (4 + 9)/2 = **6.5**.

*Tempting wrong choice: B (5) is the average of −2 and 12, but the rules explicitly forbid the extremes from being first/last.*

ANSWER 1: C

# Problem 2:

Let the tank's capacity be **C** gallons. One-third full means it holds C/3 gallons. Adding 4 gallons makes it half full:

$$\frac{C}{3} + 4 = \frac{C}{2}$$

Multiply everything by 6: 2C + 24 = 3C, so **C = 24**.

Check: 24/3 = 8 gallons, plus 4 = 12 gallons = 24/2. ✓

*Tempting wrong choice: B (12) would come from wrongly thinking the 4 gallons is one-third of the tank.*

ANSWER 2: D

# Problem 3:

Let a, b, c be the numbers of $1, $3, and $4 pairs. Then:
- a + b + c = 12 (pairs)
- 1a + 3b + 4c = 24 (dollars)

Subtract the first equation from the second: **2b + 3c = 12**.

Since b and c must each be at least 1, test values of c:
- c = 1: 2b = 9 → not a whole number
- c = 2: 2b = 6 → b = 3, so a = 12 − 2 − 3 = **7** ✓
- c = 3: 2b = 3 → not a whole number
- c = 4: 2b = 0 → b = 0, but he needs at least one of each type ✗

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 dollars, 7 + 3 + 2 = 12 pairs. ✓

ANSWER 3: D

# Problem 4:

The temperature drops from 32° to 20°, a decrease of 12°. Since the volume changes by 4 cc for every 3°:

Number of 3° steps: 12 ÷ 3 = 4 steps.

Each step reduces the volume by 4 cc, so going *down* in temperature means the earlier volume was *smaller*: the gas shrinks by 4 × 4 = 16 cc as it cools.

Volume at 20° = 24 − 16 = **8 cubic centimeters**.

*Tempting wrong choice: E (40) comes from adding 16 instead of subtracting — but heating expands the gas, so at the lower temperature the volume must be less than 24.*

ANSWER 4: A

# Problem 5:

The first number is 3 · (10101…01) and the second is 5 · (10101…01), each 99 digits long. To find the **last few digits of a product**, only the last few digits of the factors matter.

The first number ends in the digits …0303, and the second ends in …0505. So multiply:

$$303 \times 505 = 153{,}015$$

The product therefore ends in the four digits **3015**.

- Units digit: **B = 5**
- Thousands digit: **A = 3**

A + B = 3 + 5 = **8**.

(One can verify this reasoning with the full pattern: each factor equals its last 6 digits plus multiples of 10⁶, so the product's last digits match 030303 × 050505, whose last four digits are indeed 3015.)

ANSWER 5: D

# Problem 6:

To get a **large positive quotient**, either divide a big positive by a small positive, or a big negative by a small negative (negative ÷ negative = positive).

Positive ÷ positive: the best is 8 ÷ 1 = 8 (and 8 ÷ 2 = 4).

Negative ÷ negative: 
- −24 ÷ −2 = **12** ← largest so far
- −24 ÷ −3 = 8
- −3 ÷ −2 = 1.5

Any quotient involving mixed signs is negative (e.g., 8 ÷ −2 = −4), which can't be the largest.

The largest possible quotient is **12**.

ANSWER 6: D

# Problem 7:

**June:** a 20% raise multiplies her salary by 1.20:
$$2000 \times 1.20 = 2400 \text{ dollars}$$

**July:** a 20% pay cut multiplies by 0.80 — but 20% of her *new* salary:
$$2400 \times 0.80 = 1920 \text{ dollars}$$

*Tempting wrong choice: C (2000) — the raise and cut don't cancel, because the 20% cut is taken from a larger amount than the 20% raise was added to. Overall factor: 1.2 × 0.8 = 0.96, so she ends at 96% of $2000.*

ANSWER 7: A

# Problem 8:

The large cube's surface consists of 6 faces × 9 small-square faces = **54 unit squares** total.

Each corner cube of the large cube shows **3** of its faces on the surface (one on each of the three meeting faces). With 8 black corner cubes, the black surface area is 8 × 3 = **24 unit squares**. (Equivalently: each of the 6 faces has 4 black corner squares: 6 × 4 = 24.)

So the white surface area is 54 − 24 = 30 unit squares.

$$\text{Fraction white} = \frac{30}{54} = \frac{5}{9}$$

ANSWER 8: D

# Problem 9:

Let **p** = the number of games played before district play. Then:
- Wins before district play: 0.45p
- Total games for the season: p + 8 (6 wins + 2 losses)
- Total wins: 0.45p + 6

Since they finished having won half their games:

$$0.45p + 6 = \frac{p + 8}{2}$$

Multiply by 2: 0.9p + 12 = p + 8, so 4 = 0.1p, giving **p = 40**.

Total games played: 40 + 8 = **48**.

Check: 45% of 40 = 18 wins; 18 + 6 = 24 wins out of 48 games = exactly half. ✓

ANSWER 9: A

# Problem 10:

**Colfax:** 270 students in ratio 5 : 4 (9 parts). Each part = 270 ÷ 9 = 30. So boys = 150, girls = **120**.

**Winthrop:** 180 students in ratio 4 : 5 (9 parts). Each part = 180 ÷ 9 = 20. So boys = 80, girls = **100**.

**At the dance:**
- Total students: 270 + 180 = 450
- Total girls: 120 + 100 = 220

$$\text{Fraction of girls} = \frac{220}{450} = \frac{22}{45}$$

*Tempting wrong choice: D (1/2) — the ratios "look" symmetric (5:4 vs 4:5), but the schools have different sizes, so the counts don't split evenly.*

ANSWER 10: C

Problem 1:
We need the average of the first and last numbers after rearranging \(-2,4,6,9,12\).

The median is \(6\). Now check where each constrained number can go:

- The largest number, \(12\), is in one of the first three places but not first, so it must be in position 2 or 3. Thus \(12\) cannot be first or last.
- The smallest number, \(-2\), is in one of the last three places but not last, so it must be in position 3 or 4. Thus \(-2\) cannot be first or last.
- The median, \(6\), is not first or last, so it must be in position 2, 3, or 4. Thus \(6\) cannot be first or last.

So the first and last positions must be filled by the two remaining numbers, \(4\) and \(9\), in some order.

Therefore the average of the first and last numbers is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

ANSWER 1: C

Problem 2:
Let the tank capacity be \(C\) gallons.

Initially the tank is \(\frac13\) full, so it contains \(\frac13C\) gallons. After adding 4 gallons, it is \(\frac12\) full:

\[
\frac13C+4=\frac12C.
\]

Subtract \(\frac13C\) from both sides:

\[
4=\frac12C-\frac13C=\frac16C.
\]

So

\[
C=24.
\]

Check: \(\frac13(24)=8\), and \(8+4=12=\frac12(24)\).

ANSWER 2: D

Problem 3:
Let

\[
x=\text{number of \$1 pairs},
\]
\[
y=\text{number of \$3 pairs},
\]
\[
z=\text{number of \$4 pairs}.
\]

Ralph bought 12 pairs total, so

\[
x+y+z=12.
\]

The total cost was \$24, so

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
(x+3y+4z)-(x+y+z)=24-12,
\]

\[
2y+3z=12.
\]

Since he bought at least one pair of each type, \(y\ge1\) and \(z\ge1\). Try positive integer solutions:

- If \(z=1\), then \(2y=9\), impossible.
- If \(z=2\), then \(2y=6\), so \(y=3\).
- If \(z=3\), then \(2y=3\), impossible.
- If \(z=4\), then \(2y=0\), not allowed because \(y\ge1\).

Thus \(y=3\), \(z=2\). Then

\[
x=12-y-z=12-3-2=7.
\]

ANSWER 3: D

Problem 4:
The gas expands by 4 cubic centimeters for every \(3^\circ\) rise in temperature.

The temperature changed from \(20^\circ\) to \(32^\circ\), a rise of

\[
32-20=12^\circ.
\]

Since \(12^\circ\) is 4 groups of \(3^\circ\), the volume increase is

\[
4\cdot 4=16
\]

cubic centimeters.

The volume at \(32^\circ\) is 24 cubic centimeters, so the volume at \(20^\circ\) was

\[
24-16=8.
\]

ANSWER 4: A

Problem 5:
The first number has digits alternating \(3\) and \(0\), ending in \(3\). It can be written as

\[
3(1+100+100^2+\cdots+100^{49}).
\]

The second number has digits alternating \(5\) and \(0\), ending in \(5\). It can be written as

\[
5(1+100+100^2+\cdots+100^{49}).
\]

Let

\[
S=1+100+100^2+\cdots+100^{49}.
\]

The product is

\[
15S^2.
\]

We need the thousands digit and units digit, so we only need the product modulo \(10000\).

Since \(100^2=10000\), all terms from \(100^2\) onward are \(0\) modulo \(10000\). Thus

\[
S\equiv 1+100=101 \pmod{10000}.
\]

So

\[
S^2\equiv 101^2=10201\equiv 201 \pmod{10000}.
\]

Then

\[
15S^2\equiv 15\cdot 201=3015 \pmod{10000}.
\]

The last four digits of the product are \(3015\), so the thousands digit is \(A=3\) and the units digit is \(B=5\).

Thus

\[
A+B=3+5=8.
\]

ANSWER 5: D

Problem 6:
We want the largest possible quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

To get a large positive quotient, we either divide positive by positive or negative by negative.

Using only positive numbers, the largest quotient is

\[
\frac{8}{1}=8.
\]

Using two negative numbers, the quotient is positive and equals the quotient of their absolute values. To make it largest, use the largest absolute numerator and the smallest absolute denominator:

\[
\frac{-24}{-2}=12.
\]

Negative quotients are smaller than positive ones, and \(12>8\).

ANSWER 6: D

Problem 7:
Ana’s salary in May was \$2000.

In June, she received a 20% raise:

\[
2000\cdot 1.20=2400.
\]

In July, she received a 20% pay cut:

\[
2400\cdot 0.80=1920.
\]

So after both changes, her salary was \$1920.

ANSWER 7: A

Problem 8:
The large cube has side length 3, so its surface area is

\[
6\cdot 3^2=54
\]

square inches.

There are 8 corner cubes, and all 8 black cubes are placed at the corners. Each corner cube contributes 3 faces to the surface of the large cube, so the black surface area is

\[
8\cdot 3=24
\]

square inches.

Therefore the white surface area is

\[
54-24=30
\]

square inches.

The fraction of the surface area that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

ANSWER 8: D

Problem 9:
Let \(x\) be the number of games played before district play.

Before district play, the Unicorns had won 45%, so their number of wins was

\[
0.45x.
\]

During district play, they won 6 and lost 2, so they played 8 more games and had

\[
0.45x+6
\]

wins.

At the end of the season, they had won half of all games:

\[
0.45x+6=0.5(x+8).
\]

Solve:

\[
0.45x+6=0.5x+4,
\]

\[
2=0.05x,
\]

\[
x=40.
\]

So the total number of games played was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:
At Colfax Middle School, the ratio of boys to girls is \(5:4\), so girls are

\[
\frac{4}{9}
\]

of the students. With 270 students, the number of girls is

\[
\frac{4}{9}\cdot 270=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\), so girls are

\[
\frac{5}{9}
\]

of the students. With 180 students, the number of girls is

\[
\frac{5}{9}\cdot 180=100.
\]

Total girls at the dance:

\[
120+100=220.
\]

Total students at the dance:

\[
270+180=450.
\]

The fraction of students who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

ANSWER 10: C

# Problem 1:
Numbers: −2, 4, 6, 9, 12. Largest = 12, smallest = −2, median = 6.

**Constraints:**
- 12: not position 1, but within positions 1–3 → position 2 or 3.
- −2: not position 5, but within positions 3–5 → position 3 or 4.
- 6: not position 1 or 5 → position 2, 3, or 4.

**Case 12 in position 3:** then −2 must be in position 4, and 6 must go in position 2 (only remaining allowed spot). Positions 1 and 5 hold 4 and 9.

**Case 12 in position 2:** then 6 and −2 occupy positions 3 and 4 in some order, so again positions 1 and 5 hold 4 and 9.

In every case the first and last numbers are 4 and 9.

Average = (4 + 9)/2 = 6.5.

(Tempting wrong answers come from assuming the ends are e.g. −2 and 9, but −2 can't be last and 12 can't be first.)

ANSWER 1: C

# Problem 2:
Let capacity = C. Adding 4 gallons moves the level from C/3 to C/2:

C/2 − C/3 = 4 → C/6 = 4 → C = 24.

Check: one-third of 24 = 8; 8 + 4 = 12 = half of 24. ✓ (Choice B, 12, is the *amount* of water, not capacity.)

ANSWER 2: D

# Problem 3:
Let a, b, c = pairs at $1, $3, $4, each at least 1.

- a + b + c = 12
- a + 3b + 4c = 24

Subtract: 2b + 3c = 12. So 3c must be even → c is even. With c ≥ 1: c = 2 gives 2b = 6, b = 3, then a = 12 − 3 − 2 = 7. (c = 4 gives b = 0, not allowed.)

Check: 7(1) + 3(3) + 2(4) = 7 + 9 + 8 = 24 ✓ and 7 + 3 + 2 = 12 ✓

ANSWER 3: D

# Problem 4:
Temperature dropped from 32° to 20°, a drop of 12° = 4 steps of 3°.

Each 3° drop shrinks the volume by 4 cm³, so the volume decreased by 4 × 4 = 16 cm³.

Volume at 20° = 24 − 16 = 8 cm³.

(Choice E, 40, comes from adding instead of subtracting.)

ANSWER 4: A

# Problem 5:
Both numbers have 99 digits: 3030…0303 and 5050…0505 (each ends in “03” and “05”).

To find the last four digits of the product, work mod 10000, using only each number's last four digits: 0303 and 0505.

303 × 505 = 153,015.

Last four digits: 3015. So thousands digit A = 3, units digit B = 5.

A + B = 3 + 5 = 8.

ANSWER 5: D

# Problem 6:
To make a quotient large and positive, divide a large-magnitude number by a small-magnitude number of the same sign.

- (−24)/(−2) = 12
- (−24)/(−3) = 8
- 8/1 = 8, 8/2 = 4
- (−24)/1 = −24 (negative)

The largest is 12. (24 would require dividing −24 by −1, but −1 isn't in the set.)

ANSWER 6: D

# Problem 7:
June: 2000 × 1.20 = 2400.
July: 2400 × 0.80 = 1920.

A 20% cut is taken off the *larger* amount, so it doesn't undo the 20% raise — the salary ends below $2000.

ANSWER 7: A

# Problem 8:
The big cube's surface area = 6 faces × 9 unit squares = 54 unit squares.

Each corner cube of a cube shows exactly 3 of its faces on the surface. With 8 black corner cubes: 8 × 3 = 24 black unit squares.

White squares = 54 − 24 = 30, so the white fraction = 30/54 = 5/9.

(19/27 wrongly uses counts of cubes, not surface squares.)

ANSWER 8: D

# Problem 9:
Let n = games before district play, so wins = 0.45n.

After district play: total games = n + 8, wins = 0.45n + 6, and wins = half of total:

0.45n + 6 = (n + 8)/2
0.9n + 12 = n + 8
0.1n = 4 → n = 40

Total games = 40 + 8 = 48. (Check: 18 wins before, 24 wins of 48 total = half ✓)

ANSWER 9: A

# Problem 10:
Colfax: 270 students in ratio 5:4 → 9 parts of 30, girls = 4 × 30 = 120.
Winthrop: 180 students in ratio 4:5 → 9 parts of 20, girls = 5 × 20 = 100.

Total girls = 220; total students = 270 + 180 = 450.

Fraction = 220/450 = 22/45.

(1/2 would be the answer only if the two schools were the same size.)

ANSWER 10: C

Problem 1:

We need the average of the first and last numbers after arranging  
\(-2,4,6,9,12\).

- The largest number is \(12\), and it must be in position 2 or 3.
- The smallest number is \(-2\), and it must be in position 3 or 4.
- The median is \(6\), and it cannot be first or last.

Consider the possible placements.

- If \(12\) is second and \(-2\) is third, then \(6\) must be fourth. The first and last numbers are \(4\) and \(9\).
- If \(12\) is second and \(-2\) is fourth, then \(6\) must be third. Again, the first and last numbers are \(4\) and \(9\).
- If \(12\) is third, then \(-2\) must be fourth and \(6\) must be second. Again, the first and last numbers are \(4\) and \(9\).

Thus the average is

\[
\frac{4+9}{2}=\frac{13}{2}=6.5.
\]

The other choices do not equal the forced average of the two end numbers.

ANSWER 1: C

Problem 2:

The tank is initially \(\frac13\) full and becomes \(\frac12\) full after adding 4 gallons.

The added amount is therefore

\[
\frac12-\frac13=\frac16
\]

of the tank's capacity.

So,

\[
\frac16\text{ of the capacity}=4.
\]

Multiply by 6:

\[
\text{capacity}=4\cdot 6=24.
\]

Choices such as 12 or 48 would make 4 gallons equal to \(\frac13\) or \(\frac1{12}\) of the tank, not \(\frac16\).

ANSWER 2: D

Problem 3:

Let

- \(x\) = number of pairs costing \$1,
- \(y\) = number of pairs costing \$3,
- \(z\) = number of pairs costing \$4.

There are 12 pairs total:

\[
x+y+z=12.
\]

The total cost is \$24:

\[
x+3y+4z=24.
\]

Subtract the first equation from the second:

\[
2y+3z=12.
\]

Since Ralph bought at least one of each type, \(y,z\ge1\).

Try possible positive values of \(z\). Since \(2y+3z=12\), \(z\) must be even.

If \(z=2\), then

\[
2y+3(2)=12
\]

\[
2y+6=12
\]

\[
2y=6,
\]

so \(y=3\).

Then

\[
x=12-y-z=12-3-2=7.
\]

If \(z=4\), then \(y=0\), but this is not allowed because he bought at least one pair of \$3 socks.

Therefore Ralph bought 7 pairs of \$1 socks.

ANSWER 3: D

Problem 4:

The temperature changed from \(32^\circ\) to \(20^\circ\), a decrease of

\[
32-20=12^\circ.
\]

For every \(3^\circ\) rise, the gas expands by 4 cubic centimeters. Therefore, for every \(3^\circ\) decrease, it shrinks by 4 cubic centimeters.

Since

\[
12^\circ \div 3^\circ=4,
\]

the volume decreases by

\[
4\cdot 4=16
\]

cubic centimeters.

Starting from 24 cubic centimeters:

\[
24-16=8.
\]

The answer is not 40, because 40 would be the volume after a \(12^\circ\) increase rather than a decrease.

ANSWER 4: A

Problem 5:

We need only the thousands digit and units digit of the product.

The first number ends in \(303\), and the second number ends in \(505\). To find the last four digits of the product, calculate

\[
303\cdot 505.
\]

\[
303\cdot 505=303(500+5)
\]

\[
=151500+1515
\]

\[
=153015.
\]

The last four digits are \(3015\).

Thus:

- the thousands digit is \(A=3\),
- the units digit is \(B=5\).

So,

\[
A+B=3+5=8.
\]

Choices such as 3 or 5 use only one of the requested digits rather than their sum.

ANSWER 5: D

Problem 6:

We want the largest quotient using two numbers from

\[
\{-24,-3,-2,1,2,8\}.
\]

A large positive quotient can result from dividing two negative numbers or two positive numbers.

Possible large positive quotients include:

\[
\frac{-24}{-2}=12,
\]

\[
\frac{-24}{-3}=8,
\]

\[
\frac{8}{1}=8,
\]

\[
\frac{2}{1}=2.
\]

The largest is

\[
12.
\]

A quotient such as \(\frac{-24}{1}=-24\) is negative, so it cannot be the largest when positive quotients are available. Also, 24 cannot be formed from any allowed division.

ANSWER 6: D

Problem 7:

Ana starts with a monthly salary of \$2000.

In June, she receives a 20% raise:

\[
2000(1.20)=2400.
\]

In July, she receives a 20% pay cut from the new salary:

\[
2400(0.80)=1920.
\]

Thus her salary after both changes is

\[
\$1920.
\]

It is not \$2000 because a 20% increase and a 20% decrease do not cancel: the decrease is taken from the larger amount, \$2400.

ANSWER 7: A

Problem 8:

The large cube has edge length 3 inches, so each face has area

\[
3\cdot 3=9
\]

square inches.

Since a cube has 6 faces, its total surface area is

\[
6\cdot 9=54
\]

square inches.

Each black small cube is at a corner of the large cube. A corner cube has 3 exposed faces, so each black cube contributes 3 square inches of black surface.

There are 8 black corner cubes, so the black surface area is

\[
8\cdot 3=24.
\]

Therefore the white surface area is

\[
54-24=30.
\]

The fraction that is white is

\[
\frac{30}{54}=\frac{5}{9}.
\]

The fraction \(\frac{19}{27}\) counts white small cubes, but the question asks about surface area, not the fraction of small cubes that are white.

ANSWER 8: D

Problem 9:

Let \(n\) be the number of games the Unicorns played before district play.

Before district play, they had won 45% of their games:

\[
0.45n.
\]

During district play, they won 6 games and lost 2 games. Thus:

- total games at the end: \(n+8\),
- total wins at the end: \(0.45n+6\).

They finished with half of their games won, so

\[
0.45n+6=\frac12(n+8).
\]

\[
0.45n+6=0.5n+4.
\]

Subtract \(0.45n\) and 4 from both sides:

\[
2=0.05n.
\]

\[
n=40.
\]

They then played 8 more district games, so their season total was

\[
40+8=48.
\]

ANSWER 9: A

Problem 10:

At Colfax Middle School, the ratio of boys to girls is \(5:4\). There are

\[
5+4=9
\]

total ratio parts, and there are 270 students.

Each part is

\[
270\div 9=30.
\]

So the number of girls at Colfax is

\[
4\cdot 30=120.
\]

At Winthrop Middle School, the ratio of boys to girls is \(4:5\). There are

\[
4+5=9
\]

total ratio parts, and there are 180 students.

Each part is

\[
180\div 9=20.
\]

So the number of girls at Winthrop is

\[
5\cdot 20=100.
\]

Thus, the total number of girls at the dance is

\[
120+100=220.
\]

The total number of students is

\[
270+180=450.
\]

Therefore the fraction who are girls is

\[
\frac{220}{450}=\frac{22}{45}.
\]

The answer is not \(\frac12\), since there are 220 girls and 230 boys, so girls make