🔬 Model Lab

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C

Problem 1:
We are asked for Zoe's overall percent correct, given data on Chloe and on Zoe's alone work.

Let the total number of problems be $N$.
Each girl solved $N/2$ alone and $N/2$ together with the other.

For Chloe:
Correct alone $=80\%$ of $N/2 = 0.80 \cdot N/2 = 0.40N$.
Correct overall $=88\%$ of $N = 0.88N$.
So correct on the together-half $=0.88N-0.40N=0.48N$.

The together-half has $N/2=0.50N$ problems. So the together accuracy is
$$0.48N / 0.50N = 0.96 = 96\%.$$

Chloe and Zoe solved the together-half jointly, so they have the same answers and same accuracy there. Thus Zoe also got $0.48N$ correct on the together-half.

For Zoe alone:
Correct alone $=90\%$ of $N/2 = 0.90\cdot N/2 = 0.45N$.

So Zoe's total correct $=0.45N+0.48N=0.93N$, i.e. $93\%$.

Tempting errors are to average $80\%$ and $96\%$ incorrectly, or to assume Zoe's together rate equals Chloe's alone rate.
ANSWER 1: C

Problem 2:
We are asked to evaluate $(3 \star 5)\star 8$ where $A\star B=(A+B)/2$.

First inner operation:
$$3\star 5 = \frac{3+5}{2}=\frac{8}{2}=4.$$

Then:
$$(3\star 5)\star 8 = 4\star 8 = \frac{4+8}{2}=\frac{12}{2}=6.$$

This is the average operation; do not confuse with sum or product. $30$, $16$, $12$, $8$ come from mis-applying the definition.
ANSWER 2: A

Problem 3:
Each cube has 1 protruding male snap and 5 female holes. Snapping hides 1 male in 1 female. We want no male showing; all exposed faces must be female.

Let $n$ be number of cubes. There are $n$ males total. Each snap connection hides exactly 1 male. So to hide all $n$ males we need at least $n$ connections.

For $n=3$: with cubes on a square grid, any connected set of 3 cubes has at most 2 face-adjacencies, i.e. at most 2 connections, $<3$. So at least one male remains exposed. Also 3 grid cells cannot form a cycle; the smallest cycle is a $2\times2$ square.

For $n=4$: arrange in a $2\times2$ square loop. Each cube is adjacent to two neighbours, giving 4 connections total. Orient each cube's male clockwise into the next cube's female. Then each cube uses its male for one connection and a female for the other connection, leaving 4 exposed females per cube. All males hidden, only females show.

Hence 3 is impossible and 4 is achievable. This is minimal.
ANSWER 3: B

Problem 4:
We ask in how many ways $47$ is the sum of two primes.

$47$ is odd. The sum of two integers is odd only if one is even and one is odd.

The only even prime is $2$. So any representation as sum of two primes must be $47=2+p$ with $p$ prime.

Then $p=47-2=45$, which is not prime; it is $5\times 9$.

The sum of two odd primes is even, so cannot be $47$. Hence there is no representation.

Order does not create a new way. So $0$ ways.
ANSWER 4: A

Problem 5:
We are asked how many hours to shovel volume $4\times10\times3$.

Volume to remove:
$$V=4\cdot10\cdot3=120\text{ cubic yards}.$$

Rate hour 1: $20$, hour 2: $19$, hour 3: $18$, etc. After $n$ hours total shovelled:
$$20+19+\cdots+(21-n)=n\frac{20+(21-n)}{2}= \frac{n(41-n)}{2}.$$

Set $\approx120$:
$$n(41-n)\approx240.$$

After 6 hours: $(20+15)\cdot6/2=35\cdot3=105$.
After 7 hours: add $14$ to get $119$.

So after 7 hours almost all $120$ is done; total time $\approx7.07$ hours. Closest integer choice is $7$, not $4,5,6$ or $12$.
ANSWER 5: D

Problem 6:
We are asked for average miles per gallon $=$ total miles $/$ total gallons consumed.

Miles: $57{,}060-56{,}200=860$ miles.

Track fuel: Start by adding 6 gal to become full. Drive $56{,}560-56{,}200=360$ miles, add 12 gal to become full again. So the first leg consumed 12 gal.

Drive $57{,}060-56{,}560=500$ miles, add 20 gal to become full again. So the second leg consumed 20 gal.

Total consumed on trip $=12+20=32$ gal, since tank starts full and ends full.

Hence mpg $=860/32=26.875\approx26.9$.

The trap is to divide by $6+12+20=38$ to get $\approx22.6$; the initial 6 gal was in the tank before the trip and is not consumed unless counted with a full final tank correction. Correct accounting gives $26.9$.
ANSWER 6: D

Problem 7:
We count seatings of Ali, Bonnie, Carlo, Dianna in 4 distinct seats with driver restricted.

Seats are distinct: driver, front passenger, back-left, back-right.

Driver: only Bonnie or Carlo can drive: $2$ choices.

After driver is fixed, $3$ people remain for $3$ distinct remaining seats: $3! =6$ ways.

Total $=2\cdot6=12$.

If the two back seats were considered identical, we would get $6$, but physical seats are distinct positions.
ANSWER 7: D

Problem 8:
We are asked for the true combined discount of $30\%$ off then $20\%$ off.

Let original price be $P$.
After $30\%$ off: $P-0.30P=0.70P$.
After additional $20\%$ off the sale price: $0.70P\cdot(1-0.20)=0.70P\cdot0.80=0.56P$.

Amount paid is $56\%$ of original, so discount is $1-0.56=0.44=44\%$.

Discounts do not add: $30\%+20\%=50\%$ is false because the second percent is of the reduced price.
ANSWER 8: B

Problem 9:
We have a $4\times4\times4$ box filled with $64$ small cubes. Count cubes touching a lateral side or the bottom, i.e. not the top alone.

Let coordinates $x,y,z\in\{1,2,3,4\}$, $z=1$ bottom, $z=4$ top. Touch side means $x=1$ or $4$ or $y=1$ or $4$. Touch bottom means $z=1$.

Complement: touch neither side nor bottom means $x\in\{2,3\}$, $y\in\{2,3\}$, $z\in\{2,3,4\}$. That is $2\cdot2\cdot3=12$ cubes. These are the only cubes that avoid the 4 sides and bottom; they are interior in $x,y$ and off the bottom.

Hence number touching $=64-12=52$.

Note $48$, $60$, $64$ correspond to miscounts; including the top would give $64-8=56$, not listed.
ANSWER 9: B

Problem 10:
We must place digits $2,4,5,6,9$ in the boxes of a $3$-digit minus $2$-digit subtraction to get the smallest possible difference. Let it be $ABC-DE$.

Since smallest $ABC\ge245$ and largest $DE\le96$, $ABC>DE$ always, difference positive. To minimize $100A+10B+C-10D-E$, make $A$ minimal, then $B$ small and $D$ large, then $C$ small and $E$ large.

Take $A=2$, the smallest digit. From remaining $\{4,5,6,9\}

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

---

### Problem 1:
**Restatement:**
Chloe and Zoe each solve half the problems alone and the other half together. Chloe gets $80\%$ correct alone and $88\%$ correct overall. Zoe gets $90\%$ correct alone. We want to find Zoe's overall percentage of correct answers.

**Step-by-step Solution:**
1. Let the total number of problems on the homework assignment be $2N$.
2. Chloe solves $N$ problems alone and gets $80\%$ of them right, which is $0.80N$ correct answers.
3. Together, Chloe and Zoe solve the remaining $N$ problems. Let $C_{\text{together}}$ be the number of problems they answer correctly together.
4. Chloe's overall score is $88\%$, so:
   $$\frac{0.80N + C_{\text{together}}}{2N} = 0.88$$
   $$0.80N + C_{\text{together}} = 1.76N \implies C_{\text{together}} = 0.96N$$
   This means they solved $96\%$ of the collaborative problems correctly.
5. Zoe solves $N$ problems alone and gets $90\%$ correct, giving $0.90N$ correct answers.
6. Since they worked on the second half together, Zoe also gets the same $0.96N$ problems correct on that half.
7. Zoe's overall percentage of correct answers is:
   $$\frac{0.90N + 0.96N}{2N} = \frac{1.86N}{2N} = 0.93 = 93\%$$

ANSWER 1: C

---

### Problem 2:
**Restatement:**
Given the operation $A \ast B = \frac{A + B}{2}$, find the value of $(3 \ast 5) \ast 8$.

**Step-by-step Solution:**
1. First, evaluate the expression inside the parentheses:
   $$3 \ast 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$
2. Now, evaluate the outer operation using the result:
   $$4 \ast 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 2: A

---

### Problem 3:
**Restatement:**
Each cube has 1 snap on one face and receptacle holes on the other 5 faces. We want to find the minimum number of cubes snapped together such that no snaps are visible (every snap is inserted into a receptacle hole).

**Step-by-step Solution:**
1. Each cube has exactly one snap. For no snaps to show, every cube's snap must be inserted into a receptacle hole of an adjacent cube.
2. Two cubes cannot hide each other's snaps because a snap face cannot receive another snap face (the snaps would collide). Thus, at least 3 cubes are needed.
3. In a 3D cubic grid, any closed cycle of face-to-face connected cubes must have an even number of steps because the grid is bipartite. Therefore, a cycle of 3 cubes is impossible.
4. A cycle of 4 cubes arranged in a $2 \times 2$ square loop is possible:
   - Cube 1 snaps into Cube 2,
   - Cube 2 snaps into Cube 3,
   - Cube 3 snaps into Cube 4,
   - Cube 4 snaps into Cube 1.
5. In this configuration, all 4 snaps are plugged into receptacle holes, leaving only receptacle holes visible on the exterior. Thus, the minimum number is 4.

ANSWER 3: B

---

### Problem 4:
**Restatement:**
Find the number of ways to write 47 as the sum of two prime numbers.

**Step-by-step Solution:**
1. Suppose $47 = p + q$, where $p$ and $q$ are prime numbers.
2. The number 47 is odd. The sum of two integers is odd if and only if one integer is even and the other is odd.
3. The only even prime number is 2. Therefore, one of the primes must be 2.
4. This requires the other prime to be:
   $$47 - 2 = 45$$
5. However, $45$ is composite ($45 = 3^2 \times 5$).
6. Thus, 47 cannot be represented as the sum of two primes, giving 0 ways.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Pauline shovels snow from a driveway that is 4 yards wide, 10 yards long, and 3 yards deep. Her shoveling rate starts at 20 cubic yards for the first hour and decreases by 1 cubic yard each subsequent hour. Find the closest integer number of hours it will take her to shovel the driveway completely.

**Step-by-step Solution:**
1. Calculate the total volume of snow:
   $$V = 4 \times 10 \times 3 = 120\text{ cubic yards}$$
2. The amount of snow Pauline shovels in the first $n$ hours is the sum of the arithmetic series:
   $$S_n = 20 + 19 + 18 + \dots + (21 - n) = \frac{n(20 + (21 - n))}{2} = \frac{n(41 - n)}{2}$$
3. Evaluate $S_n$ for various values of $n$:
   - For $n = 6$: $S_6 = \frac{6 \times 35}{2} = 105\text{ cubic yards}$
   - For $n = 7$: $S_7 = \frac{7 \times 34}{2} = 119\text{ cubic yards}$
   - For $n = 8$: $S_8 = \frac{8 \times 33}{2} = 132\text{ cubic yards}$
4. After 7 hours, she has shoveled 119 cubic yards, leaving just 1 cubic yard. In the 8th hour, she shovels at a rate of 13 cubic yards per hour, so the remaining 1 cubic yard takes $\frac{1}{13}$ of an hour ($\approx 4.6$ minutes).
5. The total time taken is approximately $7.08$ hours, which is closest to 7.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
The odometer begins at 56,200 miles and the tank is filled with 6 gallons. At 56,560 miles, 12 gallons are added. At the end, at 57,060 miles, 20 gallons are added to fill the tank again. Find the average miles-per-gallon (mpg) for the trip to the nearest tenth.

**Step-by-step Solution:**
1. The total distance driven during the trip is:
   $$\text{Distance} = 57,060 - 56,200 = 860\text{ miles}$$
2. The 6 gallons purchased at the very start were needed to fill the tank initially (fuel consumed before the trip began). 
3. The total amount of fuel actually consumed during the trip is the fuel needed to replenish what was burned, which is the 12 gallons added mid-trip plus the 20 gallons added at the end of the trip:
   $$\text{Fuel used} = 12 + 20 = 32\text{ gallons}$$
4. Calculate the average miles per gallon:
   $$\text{mpg} = \frac{860}{32} = \frac{215}{8} = 26.875 \approx 26.9\text{ mpg}$$

ANSWER 6: D

---

### Problem 7:
**Restatement:**
Four people (Ali, Bonnie, Carlo, Dianna) are seated in a car with 4 distinct seats (driver, front passenger, two back seats). Only Bonnie and Carlo can drive. Find the number of possible seating arrangements.

**Step-by-step Solution:**
1. The driver's seat must be occupied by someone who can drive. There are 2 choices (Bonnie or Carlo).
2. Once the driver is selected, there are 3 remaining passengers to be seat

# Solutions

## Problem 1:
Let the total assignment have 2n problems (each girl does n alone, then n together).

**Chloe:** She got 80% of her n solo problems right = 0.8n. Overall she got 88% of 2n = 1.76n right. So on the n together problems she got 1.76n − 0.8n = 0.96n right — that is, **96% of the together problems were answered correctly**.

**Zoe:** The together problems are the same, so she also got 96% of those n problems right. Her solo percentage was 90%.
- Correct answers: 0.90n + 0.96n = 1.86n out of 2n total.
- Overall percentage: 1.86n / 2n = 93%.

(Checking wrong choices: 89 would require the shared portion to be 88%; 98 requires it to exceed 100% — impossible.)

ANSWER 1: C

## Problem 2:
The operation is A ✶ B = (A+B)/2 (the average).

Step 1: Compute the parentheses first:
3 ✶ 5 = (3+5)/2 = 8/2 = 4

Step 2: Now compute 4 ✶ 8:
4 ✶ 8 = (4+8)/2 = 12/2 = 6

(Tempting error: computing (3+5+8)/2 = 8, which ignores that the operation is binary and applies to only two numbers at a time.)

ANSWER 2: A

## Problem 3:
Each cube has 1 snap that must be hidden inside a receptacle of another cube. With n cubes, we need n snap-to-receptacle connections, and each connection glues two faces together.

**Try n = 3:** A structure of 3 cubes face-to-face is either a straight line or an L-shape (a "tree"). A tree of 3 cubes has only 2 connections — one snap would remain exposed. A 3-cycle is impossible with cubes face-to-face. So 3 fails.

**Try n = 4:** Arrange 4 cubes in a 2×2 square. This has 4 face-to-face connections (a cycle: cube 1–2, 2–3, 3–4, 4–1). Each cube touches two neighbors: it inserts its one snap into one neighbor's receptacle, and its other face receives the neighbor's snap in a receptacle. All 4 snaps are hidden, and every exposed face is a receptacle. ✓

So the smallest number is 4.

ANSWER 3: B

## Problem 4:
47 is odd. If 47 = p + q with both primes, then one of them must be even (since odd + odd = even and even + even = even). The only even prime is 2.

So we'd need 47 = 2 + 45. But 45 = 9 × 5 is not prime.

Therefore 47 **cannot** be written as a sum of two primes — there are 0 ways.

ANSWER 4: A

## Problem 5:
Volume of snow: 4 × 10 × 3 = 120 cubic yards.

Pauline's cumulative shoveling by hour:
- Hour 1: 20 (total 20)
- Hour 2: 19 (total 39)
- Hour 3: 18 (total 57)
- Hour 4: 17 (total 74)
- Hour 5: 16 (total 90)
- Hour 6: 15 (total 105)
- Hour 7: 14 (total 119)
- Hour 8: 13 (total 132)

Since 120 falls between 119 (7 hours) and 132 (8 hours), and 120 − 119 = 1 while 132 − 120 = 12, the time is just barely over 7 hours — closest to 7.

ANSWER 5: D

## Problem 6:
Total gasoline used over the whole trip: 6 + 12 + 20 = 38 gallons. (All fill-ups during/at the end of the trip count as gas used on the trip; the tank is filled back up at the end.)

Total miles driven: 57,060 − 56,200 = 860 miles.

Average: 860 / 38 ≈ 22.63 → 22.6 mpg.

(Tempting error: using only the middle/last segment's gas, or only the last odometer reading's gap — those give other listed wrong answers like 24.0 or 26.9.)

ANSWER 6: B

## Problem 7:
**Driver's seat:** Only Bonnie and Carlo can drive → 2 choices.

**Remaining 3 seats** (front passenger + 2 back): the remaining 3 people (Ali, plus whichever of Bonnie/Carlo isn't driving) can be arranged in 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Tempting error: 24 = 4! assumes anyone can drive; 6 forgets the two driver choices.)

ANSWER 7: D

## Problem 8:
Let the original price be 100.

- After 30% off: price = 70.
- After an additional 20% off: 70 × 0.80 = 56.

Final price is 56, so the total discount is 100 − 56 = 44%.

(Tempting error: adding 30% + 20% = 50% — but the second discount is taken on the already-reduced price, not the original.)

ANSWER 8: B

## Problem 9:
The box is 4×4×4 = 64 small cubes.

**Cubes touching a side (any of the 4 walls):** The interior cubes not touching any wall form a 2×2 column arrangement in the middle (a 2×2 cross-section, 4 tall) = 4 × 4 = 16 cubes. So cubes touching a wall = 64 − 16 = 48.

**Cubes touching the bottom:** the bottom layer is a 4×4 = 16 cube layer.

**Key observation:** every cube in the bottom layer lies on the 4×4 base, so all of them touch a wall of the box as well. Therefore the bottom layer adds **no new cubes** beyond those already counted.

Total = 48.

(Tempting error: 48 + 16 = 64, double-counting the bottom layer.)

ANSWER 9: A

## Problem 10:
The five boxes form a three-digit number minus a two-digit number (5 digits total), using digits 2, 4, 5, 6, 9, and we minimize the difference.

To make the difference small:
- Make the three-digit number as **small** as possible → hundreds digit = 2.
- Make the two-digit number as **large** as possible → use 9 in the tens place.

So we need the smallest possible value of (2 b c) − (9 d e) using remaining digits {4, 5, 6}:
- Difference = 200 + 10b + c − 90 − d = 110 + (10b + c − 10d − e)... let's just compute candidates:
- 245 − 96 = 149
- 246 − 95 = 151
- 254 − 96 = 158
- 256 − 94 = 162, 265 − 94 = 171, etc. — all larger.

The minimum is 245 − 96 = **149**. (Any other hundreds digit makes the three-digit number at least 400, giving a larger difference; any smaller subtrahend gives a larger difference too.)

ANSWER 10: C

Problem 1:

Chloe solved half the problems alone and half together with Zoe.

Let the percentage correct on the problems they solved together be \(x\%\).

Chloe’s overall percentage was \(88\%\), so
\[
\frac{80+x}{2}=88.
\]
Thus
\[
80+x=176,
\]
so
\[
x=96.
\]

Zoe got \(90\%\) correct on her half solved alone and \(96\%\) correct on the half solved together. Therefore her overall percentage was
\[
\frac{90+96}{2}=93.
\]

The tempting answer \(92\%\) would result from incorrectly averaging \(88\%\) and \(96\%\), rather than using Zoe’s own \(90\%\) solo score.

ANSWER 1: C

Problem 2:

The symbol \(\star\) means “take the average”:
\[
A\star B=\frac{A+B}{2}.
\]

First calculate the expression inside the parentheses:
\[
3\star 5=\frac{3+5}{2}=4.
\]

Now calculate
\[
4\star 8=\frac{4+8}{2}=6.
\]

A common mistake is to average all three numbers at once, but the parentheses tell us to do \(3\star5\) first.

ANSWER 2: A

Problem 3:

Each cube has exactly one protruding snap. For only receptacle holes to show, every protruding snap must be attached to a receptacle hole of another cube.

Thus, the cubes must form a closed loop: each cube’s protruding snap attaches to the next cube.

A loop of 3 cubes cannot be made from cubes arranged face-to-face, since face-to-face cube connections lie on a grid and cannot form a triangle. The smallest possible loop is a square of 4 cubes.

Arrange 4 cubes in a square, with each cube’s protruding snap connected to the next cube. Then all four protruding snaps are hidden, and all exposed faces have receptacle holes.

Thus the smallest number is \(4\).

The answer is not 3 because three cubes cannot make a face-to-face closed cycle. Larger choices work in some arrangements but are not smallest.

ANSWER 3: B

Problem 4:

Since \(47\) is odd, a sum of two primes can equal \(47\) only if one of the primes is \(2\).

This is because every prime except \(2\) is odd, and
\[
\text{odd}+\text{odd}=\text{even}.
\]

So we would need
\[
47=2+45.
\]

But \(45\) is not prime:
\[
45=3\cdot 15.
\]

Therefore, there are no ways to write \(47\) as the sum of two primes.

ANSWER 4: A

Problem 5:

The driveway’s volume is
\[
4\text{ yards}\times 10\text{ yards}\times 3\text{ yards}=120\text{ cubic yards}.
\]

Pauline shovels the following amounts:

\[
\begin{array}{c|c|c}
\text{Hour} & \text{Shoveled that hour} & \text{Total shoveled}\\
\hline
1&20&20\\
2&19&39\\
3&18&57\\
4&17&74\\
5&16&90\\
6&15&105\\
7&14&119
\end{array}
\]

After 7 hours, she has shoveled \(119\) cubic yards, leaving only
\[
120-119=1
\]
cubic yard.

During hour 8, she shovels at a rate of \(13\) cubic yards per hour, so the remaining work takes
\[
\frac{1}{13}
\]
of an hour.

Thus the total time is
\[
7+\frac{1}{13}\approx 7.1,
\]
which is closest to \(7\).

The answer is not \(6\), since only \(105\) cubic yards have been cleared after 6 hours.

ANSWER 5: D

Problem 6:

The total distance traveled was
\[
57,\!060-56,\!200=860\text{ miles}.
\]

The initial 6 gallons were added before the trip began, so they should not be counted as gasoline used during the trip.

During the trip, the driver used the gasoline that was replaced by the later fill-ups:
\[
12+20=32\text{ gallons}.
\]

Therefore the average miles per gallon was
\[
\frac{860}{32}=26.875.
\]

To the nearest tenth,
\[
26.875\approx 26.9.
\]

A tempting wrong answer is \(22.6\), which comes from incorrectly including the initial 6 gallons:
\[
\frac{860}{6+12+20}\approx22.6.
\]

ANSWER 6: D

Problem 7:

Bonnie and Carlo are the only two people who can drive, so there are

\[
2
\]
choices for the driver’s seat.

After choosing the driver, the remaining 3 people can be arranged in the remaining 3 seats in
\[
3!=3\cdot2\cdot1=6
\]
ways.

Therefore the total number of seating arrangements is
\[
2\cdot 6=12.
\]

The answer is not \(24\), since not all four people are eligible to sit in the driver’s seat.

ANSWER 7: D

Problem 8:

Suppose the original price is \(\$100\).

After a \(30\%\) discount, the price is
\[
100(1-0.30)=70.
\]

Then the merchant takes \(20\%\) off the sale price:
\[
70(1-0.20)=70(0.80)=56.
\]

The final price is \(\$56\), so the discount from the original \(\$100\) price is
\[
100-56=44.
\]

Thus the total discount is \(44\%\).

It is incorrect to add \(30\%\) and \(20\%\) to get \(50\%\), because the \(20\%\) discount is taken from the already reduced price.

ANSWER 8: B

Problem 9:

There are
\[
4\cdot4\cdot4=64
\]
small cubes altogether.

It is easier to count the cubes that do **not** touch a side or the bottom.

To avoid touching a side, a cube must be in one of the middle \(2\times2\) positions horizontally:
\[
2\cdot2=4
\]
possibilities.

To avoid touching the bottom, it can be in any of the upper 3 layers:
\[
3
\]
possibilities.

So the number that touch neither a side nor the bottom is
\[
2\cdot2\cdot3=12.
\]

Therefore, the number that touch a side or the bottom is
\[
64-12=52.
\]

The answer \(48\) would miss some cubes that touch the bottom but not a side.

ANSWER 9: B

Problem 10:

The subtraction has a three-digit number minus a two-digit number, using the digits \(2,4,5,6,9\) exactly once.

To make the difference as small as possible, we want:

- the three-digit number as small as possible;
- the two-digit number as large as possible.

The smallest possible three-digit number begins with \(2\). The smallest remaining arrangement for the last two places is \(45\), giving
\[
245.
\]

The largest possible two-digit number from the other digits is
\[
96.
\]

Thus we can make
\[
245-96=149.
\]

No smaller result is possible: the three-digit number cannot be less than \(245\), and the two-digit number cannot be greater than \(96\). Therefore,
\[
245-96=149
\]
is the smallest possible difference.

Choices such as \(123\) would require a number smaller than \(245\) or a subtracted number larger than \(96\), neither of which is possible.

ANSWER 10: C