🔬 Model Lab

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D

Problem 1:
We are given that $A$, $B$, $C$, and $D$ are digits, with:
1. $AB + CA = DA$
2. $AB - CA = A$

Writing these two-digit numbers in base 10:
- $AB = 10A + B$
- $CA = 10C + A$
- $DA = 10D + A$

Adding the two given equations eliminates $CA$:
$$(AB + CA) + (AB - CA) = DA + A$$
$$2(AB) = (10D + A) + A = 10D + 2A$$
$$2(10A + B) = 10D + 2A$$
$$20A + 2B = 10D + 2A \implies 18A + 2B = 10D \implies 9A + B = 5D$$

From the second equation, $AB - CA = A$:
$$(10A + B) - (10C + A) = A \implies 9A + B - 10C = A \implies 8A + B = 10C$$

Since $C$ is a non-zero digit, $10C$ is a multiple of 10, meaning the units digit of $8A + B$ must be 0. Also, from $9A + B = 5D$, $9A + B$ must be a multiple of 5.
Subtracting the two equations:
$$(9A + B) - (8A + B) = 5D - 10C \implies A = 5(D - 2C)$$
Since $A$ is a non-zero digit and a multiple of 5, we must have $A = 5$.

Now substitute $A = 5$:
- $8(5) + B = 40 + B = 10C$. Since $B$ is a single digit ($0 \le B \le 9$), the only multiple of 10 in this range is 40, so $B = 0$ and $C = 4$.
- Then, $9A + B = 9(5) + 0 = 45 = 5D \implies D = 9$.

Let us check:
- $AB = 50$, $CA = 45$, $DA = 95$
- $50 + 45 = 95$ (correct)
- $50 - 45 = 5 = A$ (correct)

Thus, the digit $D$ represents is 9.

ANSWER 1: E

---

Problem 2:
We want to find the total distance Karl drove that day.
1. Karl starts with a full tank of 14 gallons.
2. He drives 350 miles. Since his car gets 35 miles per gallon, he uses:
   $$\frac{350\text{ miles}}{35\text{ mpg}} = 10\text{ gallons}$$
   The amount of gas remaining in the tank is:
   $$14 - 10 = 4\text{ gallons}$$
3. He then buys 8 gallons of gas, bringing his tank to:
   $$4 + 8 = 12\text{ gallons}$$
4. At his destination, the gas tank is half full, which means it contains:
   $$\frac{14}{2} = 7\text{ gallons}$$
5. The amount of gas Karl used on the second leg of his trip is:
   $$12 - 7 = 5\text{ gallons}$$
6. The distance driven on the second leg is:
   $$5\text{ gallons} \times 35\text{ mpg} = 175\text{ miles}$$

The total miles driven that day is:
$$350 + 175 = 525\text{ miles}$$

ANSWER 2: A

---

Problem 3:
Let the square be placed in the coordinate plane with corners at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$. The side length is 10 meters.
For any point $(x,y)$ strictly inside the square (where $0 \le x \le 10$ and $0 \le y \le 10$):
- The shortest distance to the left side ($x = 0$) is $x$.
- The shortest distance to the right side ($x = 10$) is $10 - x$.
- The shortest distance to the bottom side ($y = 0$) is $y$.
- The shortest distance to the top side ($y = 10$) is $10 - y$.

The sum of the distances to the four sides is:
$$x + (10 - x) + y + (10 - y) = 10 + 10 = 20\text{ meters}$$

Since the lemming starts at $(0,0)$, moves $6.2$ meters along the diagonal (reaching approximately $(4.38, 4.38)$), and then turns 90° right and runs 2 meters, it remains well inside the square.

Therefore, the sum of the four shortest distances is always 20 meters, and the average of these four distances is:
$$\frac{20}{4} = 5\text{ meters}$$

ANSWER 3: C

---

Problem 4:
Let:
- $M$ be the set of students taking a math class.
- $F$ be the set of students taking a foreign language class.

We are given:
- Total students $|M \cup F| = 93$
- Students taking math $|M| = 70$
- Students taking a foreign language $|F| = 54$

We want to find the number of students who take only a math class, which is $|M \setminus F|$.
By the principle of inclusion-exclusion:
$$|M \cup F| = |M| + |F| - |M \cap F|$$
$$93 = 70 + 54 - |M \cap F| = 124 - |M \cap F| \implies |M \cap F| = 124 - 93 = 31$$

The number of students taking only a math class is:
$$|M \setminus F| = |M| - |M \cap F| = 70 - 31 = 39$$
Alternatively, since every student is in $M \cup F$, the students taking only math are those in the class who do not take a foreign language:
$$|M \cup F| - |F| = 93 - 54 = 39$$

ANSWER 4: D

---

Problem 5:
The operation is defined as $A \mathbin{*} B = \frac{A + B}{2}$.
We evaluate the expression $(3 \mathbin{*} 5) \mathbin{*} 8$ following the order of operations:

First, evaluate the expression inside the parentheses:
$$3 \mathbin{*} 5 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$

Next, substitute this result into the outer operation:
$$4 \mathbin{*} 8 = \frac{4 + 8}{2} = \frac{12}{2} = 6$$

ANSWER 5: A

---

Problem 6:
Henry starts with 1 piece of pasta.
Each time he takes a bite from the middle of an existing piece, that piece is divided into 2 pieces, increasing the total number of pieces by 1.
Since he ends up with 10 pieces, the number of bites he took is:
$$10 - 1 = 9\text{ bites}$$

Each bite consumes 3 inches of pasta, so the total length of pasta eaten is:
$$9 \times 3 = 27\text{ inches}$$

The remaining pasta has a total length of 17 inches. Therefore, the original length was:
$$17 + 27 = 44\text{ inches}$$

ANSWER 6: D

---

Problem 7:
To calculate the average miles-per-gallon (mpg) for the entire trip:
1. At the beginning of the trip, the tank was filled to full at odometer reading 56,200. (The 6 gallons put in at this point replaced fuel used before the trip began.)
2. At the end of the trip, the odometer read 57,060, and the tank was filled back to full.
3. The total distance traveled on the trip is:
   $$57,060 - 56,200 = 860\text{ miles}$$
4. The gasoline consumed during the trip is the amount required to refill the tank back to full during and at the end of the trip:
   $$12\text{ gallons} + 20\text{ gallons} = 32\text{ gallons}$$

The average fuel economy is:
$$\text{mpg} = \frac{860\text{ miles}}{32\text{ gallons}} = 26.875\text{ mpg}$$
Rounding to the nearest tenth gives $26.9\text{ mpg}$.

ANSWER 7: D

---

Problem 8:
A year is "lucky" if there exists a valid month $m \in \{1, 2, \dots, 12\}$ and day $d \in \{1, 2, \dots, 31\}$ (matching the number of days in that month) such that:
$$m \times d = \text{last two digits of the year}$$

Let us test the choices:
- A. 1990: $m \times d = 90$. We can have October 9 ($10 \times 9 = 90$), which is valid.
- B. 1991: 

# Problem 1:
We need digits A, B, C, D with two-digit numbers AB, CA, DA.

Write AB = 10A + B, CA = 10C + A, DA = 10D + A.

**Equation 1 (sum):** 10A + B + 10C + A = 10D + A → 10A + B + 10C = 10D
**Equation 2 (difference):** (10A + B) − (10C + A) = A → 9A + B − 10C = A → 8A + B = 10C

From Equation 2: B = 10C − 8A. Substitute into Equation 1:

10A + (10C − 8A) + 10C = 10D → 2A + 20C = 10D → A + 10C = 5D

So A must be a multiple of 5, and A ≠ 0 (it leads a two-digit number), so **A = 5**.
Then 5 + 10C = 5D → D = 1 + 2C, and B = 10C − 40 forces C = 4 (so that 0 ≤ B ≤ 9), giving B = 0, D = 9.

Check: 50 + 45 = 95 ✓ (D = 9, A = 5) and 50 − 45 = 5 = A ✓.

ANSWER 1: E

# Problem 2:
Full tank = 14 gal → range 14 × 35 = 490 miles.

- Drove 350 miles: used 350/35 = 10 gallons, leaving 14 − 10 = 4 gallons.
- Bought 8 gallons: now 4 + 8 = 12 gallons.
- Arrived with tank half full = 7 gallons, so he used 12 − 7 = 5 gallons after refueling: 5 × 35 = 175 miles.

Total = 350 + 175 = **525 miles**.

(Tempting wrong answers: 560 comes from wrongly assuming the tank was filled to 14; 665 from using all 12 gallons.)

ANSWER 2: A

# Problem 3:
Put the square with corners (0,0), (10,0), (10,10), (0,10); the lemming starts at (0,0).

Key idea: for any point (x, y) inside the square, the four distances to the sides are x, 10 − x, y, 10 − y, and their sum is always
x + (10 − x) + y + (10 − y) = 20.

So the average is 20/4 = **5**, regardless of where the lemming stops (as long as it's inside).

(Check: running 6.2 m along the diagonal puts it at about (4.38, 4.38); a right turn and 2 m changes x by +√2 and y by −√2, keeping it inside. Sum still 20.)

ANSWER 3: C

# Problem 4:
By inclusion–exclusion, the number taking **both** is
70 + 54 − 93 = 31.

So the number taking **only math** = 70 − 31 = **39**.

(16 would be "only foreign language" minus... actually 54 − 31 = 23 is "only foreign language," and 31 is "both" — both are tempting distractors.)

ANSWER 4: D

# Problem 5:
3 ✶ 5 = (3 + 5)/2 = 4.
Then 4 ✶ 8 = (4 + 8)/2 = 6.

ANSWER 5: A

# Problem 6:
Each bite removes 3 inches from the middle of a piece, which splits that piece into two pieces — so each bite increases the number of pieces by exactly 1.

Starting with 1 piece and ending with 10 pieces means 9 bites, eating 9 × 3 = 27 inches.

Original length = 17 + 27 = **44 inches**.

ANSWER 6: D

# Problem 7:
Distance traveled: 57,060 − 56,200 = 860 miles.

The first 6 gallons only *filled the tank* before the trip; the gas actually consumed during the trip is what was needed to refill the tank afterward: 12 + 20 = 32 gallons.

860 ÷ 32 = 26.875 ≈ **26.9 mpg**.

(22.6 comes from mistakenly including the initial 6 gallons: 860/38 ≈ 22.6.)

ANSWER 7: D

# Problem 8:
We need month (1–12) × day (valid for that month) = last two digits.

- **90:** 9 × 10 → 9/10/90 ✓ lucky
- **91:** 91 = 7 × 13 → 7/13/91 ✓ lucky
- **92:** 92 = 4 × 23 → 4/23/92 ✓ lucky
- **93:** 93 = 3 × 31 → 3/31/93 ✓ (March has 31 days) lucky
- **94:** factor pairs are 1 × 94 and 2 × 47. Day 94 impossible; day 47 impossible. No valid date → **not lucky**

ANSWER 8: E

# Problem 9:
Place A at the center. The M's are the 4 squares adjacent to A; the C's are the next ring out; the 8's are on the outside.

By the symmetry of the diamond, count paths through one M and multiply by 4.

Take the M directly right of A. From it, you can move to 3 different C's (the one straight out, and the two diagonal-ring ones above and below it). From each of those C's, there are exactly 2 adjacent 8's.

So paths through that M: 3 × 2 = 6.

Total: 4 M's × 6 = **24**.

ANSWER 9: D

# Problem 10:
Total ways to pick 2 different numbers from 6: C(6,2) = 15.

The product is 0 exactly when 0 is one of the chosen numbers; the other can be any of the remaining 5 numbers: 5 pairs.

Probability = 5/15 = **1/3**.

ANSWER 10: D

Problem 1:

We are told that \(AB, CA,\) and \(DA\) are two-digit numbers:

\[
AB=10A+B,\qquad CA=10C+A,\qquad DA=10D+A.
\]

The subtraction condition is

\[
AB-CA=A.
\]

Substitute the two-digit forms:

\[
(10A+B)-(10C+A)=A.
\]

\[
9A+B-10C=A
\]

\[
8A+B=10C.
\]

Thus \(8A+B\) must be a multiple of \(10\). In particular, \(B\) must make the last digit of \(8A+B\) equal to \(0\).

The addition condition is

\[
AB+CA=DA.
\]

Using the subtraction fact \(AB=CA+A\),

\[
(CA+A)+CA=DA,
\]

so

\[
2CA+A=DA.
\]

Try the digit possibility that works with \(8A+B=10C\). If \(A=5\), then

\[
8A=40,
\]

so \(B=0\), giving

\[
C=\frac{8A+B}{10}=\frac{40}{10}=4.
\]

Now check:

\[
AB=50,\qquad CA=45.
\]

Indeed,

\[
50-45=5=A
\]

and

\[
50+45=95=DA.
\]

Therefore \(D=9\).

Other choices \(5,6,7,8\) do not produce two-digit numbers satisfying both equations.

ANSWER 1: E

Problem 2:

Karl’s car uses 1 gallon every 35 miles.

He begins with 14 gallons. After driving 350 miles, he has used

\[
\frac{350}{35}=10
\]

gallons.

So he has left

\[
14-10=4
\]

gallons.

He then buys 8 gallons, so he has

\[
4+8=12
\]

gallons.

At the destination, the tank is half full. Since a full tank holds 14 gallons, half full means

\[
\frac{14}{2}=7
\]

gallons remain.

Thus, after buying gas, he used

\[
12-7=5
\]

gallons. This corresponds to

\[
5\cdot35=175
\]

miles.

His total distance was

\[
350+175=525.
\]

Choices such as \(560\) would mean using 16 gallons total, which is impossible because he used only \(10+5=15\) gallons.

ANSWER 2: A

Problem 3:

We need the average of the lemming’s distances from the four sides of the square.

For any point inside a square of side length 10:

- Its distances from the left and right sides add to \(10\).
- Its distances from the bottom and top sides also add to \(10\).

Therefore, the sum of all four measured distances is always

\[
10+10=20.
\]

So their average is

\[
\frac{20}{4}=5.
\]

The details of running 6.2 meters diagonally and then 2 meters at a right angle are distractions; the average is the same for every point inside the square.

ANSWER 3: C

Problem 4:

Let \(M\) be the number taking math and \(F\) be the number taking a foreign language.

We know:

\[
M=70,\qquad F=54,
\]

and every one of the 93 students takes at least one of these classes.

Using inclusion-exclusion,

\[
M+F-(\text{both})=93.
\]

So

\[
70+54-(\text{both})=93.
\]

\[
124-(\text{both})=93.
\]

\[
\text{both}=31.
\]

The number taking only math is

\[
70-31=39.
\]

Choice \(31\) is tempting, but it represents the students taking both classes, not only math.

ANSWER 4: D

Problem 5:

The operation is defined by

\[
A\mathbin{\star}B=\frac{A+B}{2}.
\]

First calculate

\[
3\mathbin{\star}5=\frac{3+5}{2}=\frac{8}{2}=4.
\]

Then calculate

\[
4\mathbin{\star}8=\frac{4+8}{2}=\frac{12}{2}=6.
\]

A common mistake is to average all three numbers at once, but the expression must be evaluated from left to right because of the parentheses.

ANSWER 5: A

Problem 6:

Henry begins with 1 piece of pasta.

Each bite is taken from the middle of one piece, so that one piece becomes two pieces. Therefore, each bite increases the number of pieces by 1.

To get from 1 piece to 10 pieces, Henry took

\[
10-1=9
\]

bites.

Each bite eats 3 inches, so he ate

\[
9\cdot3=27
\]

inches of pasta.

At the end, the remaining 10 pieces have total length 17 inches. Therefore the original length was

\[
27+17=44
\]

inches.

Choices such as \(41\) would result from incorrectly counting only 8 bites.

ANSWER 6: D

Problem 7:

The trip begins at mileage 56,200 and ends at mileage 57,060, so the total distance is

\[
57{,}060-56{,}200=860
\]

miles.

The initial 6 gallons are already in the tank at the beginning of the trip, so they should not be counted as gas used during the trip.

During the trip, the driver refills with:

\[
12+20=32
\]

gallons.

Thus the average miles per gallon is

\[
\frac{860}{32}=26.875.
\]

Rounded to the nearest tenth,

\[
26.9.
\]

The tempting answer \(22.6\) comes from incorrectly including the initial 6 gallons, giving \(860/38\).

ANSWER 7: D

Problem 8:

A lucky year has a date for which

\[
(\text{month})(\text{day})=\text{last two digits of the year}.
\]

Check each year:

- \(1990\): \(10\cdot 9=90\), so 10/9/90 works.
- \(1991\): \(7\cdot13=91\), so 7/13/91 works.
- \(1992\): \(4\cdot23=92\), so 4/23/92 works.
- \(1993\): \(3\cdot31=93\), so 3/31/93 works.
- \(1994\): the only positive factor pairs are

\[
1\cdot94,\qquad 2\cdot47.
\]

Neither day 94 nor day 47 is a valid date.

Therefore 1994 is not lucky.

ANSWER 8: E

Problem 9:

Starting from the central \(A\), the first letter must be an adjacent \(M\). Then the next letter must be an adjacent \(C\), followed by an adjacent \(8\).

Count paths according to the first \(M\) selected.

- From the top \(M\), there are 6 possible ways to continue through a \(C\) to an \(8\).
- From the left \(M\), there are 3 possible ways.
- From the right \(M\), there are 2 possible ways.
- From the bottom \(M\), there is 1 possible way.

Therefore the total number of paths spelling AMC8 is

\[
6+3+2+1=12.
\]

Choices \(8\) and \(9\) miss some valid routes, while \(24\) and \(36\) overcount by treating non-adjacent letters as available moves.

ANSWER 9: C

Problem 10:

There are 6 numbers:

\[
\{-2,-1,0,3,4,5\}.
\]

Two different numbers are selected, so the number of possible unordered pairs is

\[
\binom{6}{2}=15.
\]

The product is 0 exactly when one selected number is \(0\).

The 0 can be paired with any of the other 5 numbers, so there are

\[
5
\]

successful pairs.

Thus the probability is

\[
\frac{5}{15}=\frac13.
\]

The answer is not \(\frac16\), since selecting 0 is not the only selection; it can be paired with any one of five other numbers.

ANSWER 10: D