🔬 Model Lab

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D

Problem 1:
We are asked for the missing fourth score when three scores are 70, 80, 90 and the average of all four is 70.

Let the missing score be $x$. The average of four numbers is their sum divided by 4.
So $(70+80+90+x)/4 = 70$.

The sum of the three known scores is $70+80+90 = 240$.
So $(240+x)/4 = 70$.
Multiply by 4: $240+x = 280$.
Thus $x = 280-240 = 40$.

Check: $70+80+90+40=280$, and $280/4=70$. Correct.
Choice B 50 would give average 72.5, C 55 gives 73.75, D 60 gives 75, E 70 gives 77.5. Only 40 works.

ANSWER 1: A

Problem 2:
Eight friends agree to split a bill equally. Judi pays nothing, so the other seven each pay $2.50 extra to make up her share. Find the total bill.

Let the total be $T$. Fair share per person is $T/8$.
Judi's fair share is $T/8$. This must be covered by the 7 others paying extra.
Extra collected $= 7 \times \$2.50 = \$17.50$.

So $T/8 = 17.50$.
Thus $T = 8 \times 17.50 = 140.00$.

Check: Fair share $=140/8=17.50$. If 7 pay $17.50+2.50=20.00$, they pay $7\times20=140$. Works.
A $120 would make fair share $15, not $17.50. Similarly others do not fit.

ANSWER 2: C

Problem 3:
Two boys and three girls sit in a row, with a boy at each end and the three girls in the middle. Count arrangements.

Positions: _ _ _ _ _. Ends must be boys, middle three must be girls.

Assign the 2 distinct boys to left end and right end: there are $2! = 2$ ways. Swapping the boys gives a different picture.

Assign the 3 distinct girls to the 3 middle seats in order: there are $3! = 3\times2\times1 = 6$ ways.

By the multiplication principle, total $= 2 \times 6 = 12$.

Choice A 2 counts only boys, C 5 is $2+3$, D 6 counts only girls. We need both, so E 12.

ANSWER 3: E

Problem 4:
Abe has 1 green, 1 red. Bob has 1 green, 1 yellow, 2 red. Each picks one at random. Find P(colors match).

Abe picks green with probability $1/2$, red with $1/2$.
Bob picks green with $1/4$, yellow with $1/4$, red with $2/4=1/2$.

Match can happen as:
Both green: $(1/2)\times(1/4)=1/8$.
Both red: $(1/2)\times(2/4)=2/8=1/4$.
Both yellow: impossible since Abe has no yellow, probability 0.

Total match probability $=1/8+2/8=3/8$.

A $1/4$ is only the red-red part, D $1/2$ ignores Bob's distribution, etc. Correct is $3/8$.

ANSWER 4: C

Problem 5:
2002 is a palindrome. Find the next palindrome year after 2002 and the product of its digits.

A 4-digit palindrome has form ABBA, reads same forward and backward.
After 2002, the next numbers of form ABBA are:
2002, then 2112, then 2222, etc.
Indeed 2002 $<2112<2222$, and there is no ABBA between 2002 and 2112 because the first two digits must go from 20 to 21.

So next is 2112.
Product of digits $=2\times1\times1\times2 =4$.

Choice A 0 would require a 0 digit, as in 2002, but 2112 has none. Others miscompute.

ANSWER 5: B

Problem 6:
12 cm represents 72 km. What does 17 cm represent?

This is proportional. Km per cm $=72/12 =6$ km/cm.
So 1 cm $\to$ 6 km.
17 cm $\to$ $17\times6 =102$ km.

Check: $12\times6=72$. Correct.
A 6 is the unit rate, not the answer. C 204 is double, E 1224 is $72\times17$.

ANSWER 6: B

Problem 7:
Need 35 cents using 5, 10, 25 cent coins. Find difference between max number of coins and min number of coins possible.

Smallest number of coins: use largest values possible.
$25+10=35$ uses 2 coins. 1 coin impossible since largest is 25 $<35$. So minimum $=2$.

Largest number of coins: use smallest values possible.
All 5-cent: $35/5=7$ coins. Since 5 is the smallest denomination, you cannot use more than 7. So maximum $=7$.

Difference $=7-2=5$.

Examples in between: $25+5+5$ (3 coins), $10+10+10+5$ (4 coins), etc. Extremes are 2 and 7.

ANSWER 7: E

Problem 8:
Numbers are 3,4,5,6,6,7,x. Mean = median = unique mode. Find x.

We have 7 numbers. Already 6 appears twice. For a unique mode to exist, the mode must be 6, because any other duplicate created by x would tie or beat it unless x=6 which would still keep mode 6.
So mode $=6$. Hence mean $=6$ and median $=6$.

Mean $=6$ gives:
$(3+4+5+6+6+7+x)/7 =6$
$(31+x)/7 =6$
$31+x =42$
$x=11$.

Check median and mode with $x=11$: sorted: 3,4,5,6,6,7,11. Middle (4th) $=6$. Mode: 6 occurs twice, all others once, so unique mode 6. All three equal 6. Works.

If x=5, we get two 5s and two 6s, no unique mode. If x=6, mean $=37/7\neq6$. If x=7, two 7s tie two 6s. So only 11 works.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars. Spends 12 on food and twice as much on rides. How much left?

Rides $=2\times12 =24$ dollars.
Total spent $=12+24=36$ dollars.
Left $=50-36=14$ dollars.

A 12 is just food, C 26 is $50-24$, D 38 is $50-12$. Must subtract both.

ANSWER 9: B

Problem 10:
Evaluate $16+8\div4-2$ using order of operations.

Division before addition/subtraction.
$8\div4 =2$.
So expression $=16+2-2 =18-2=16$.

Choice C 12 would come from doing $(16+8)\div(4-2)=12$, ignoring order. Choice E 20 from $16+8-4+...$ etc. Correct with PEMDAS is 16.

ANSWER 10: D

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

---

### Problem 1:
**Restatement:** 
We are given that four students take an exam. Three of the scores are $70$, $80$, and $90$, and the average (mean) of all four scores is $70$. We need to find the fourth score.

**Step-by-step solution:**
1. The average of four scores is calculated as:
   $$\text{Average} = \frac{\text{Sum of scores}}{4}$$
2. Given that the average is $70$, the total sum of the four scores must be:
   $$\text{Sum of scores} = 4 \times 70 = 280$$
3. Let $x$ be the fourth score. Then:
   $$70 + 80 + 90 + x = 280$$
   $$240 + x = 280$$
   $$x = 280 - 240 = 40$$

4. **Evaluating choices:** 
   - A score of $40$ brings the sum to $280$, which yields an average of $70$.
   - Any higher score (such as $50$ or $60$) would make the average greater than $70$.

ANSWER 1: A

---

### Problem 2:
**Restatement:** 
Eight friends split a bill equally. One friend (Judi) forgot her money, so each of the remaining $7$ friends paid an extra $\$2.50$ to cover her share. We want to find the total bill.

**Step-by-step solution:**
1. The amount needed to cover Judi's share was split among the $7$ friends.
2. Since each paid $\$2.50$, Judi's share of the bill was:
   $$\text{Judi's share} = 7 \times \$2.50 = \$17.50$$
3. Since all $8$ friends originally agreed to share the bill equally, each person's share was $\$17.50$.
4. Therefore, the total bill was:
   $$\text{Total bill} = 8 \times \$17.50 = \$140.00$$

ANSWER 2: C

---

### Problem 3:
**Restatement:** 
A chess team consists of $2$ boys and $3$ girls. They must sit in a row of $5$ seats such that a boy sits at each end and the $3$ girls sit in the middle three seats. We want to find the number of possible seating arrangements.

**Step-by-step solution:**
1. The seating pattern is: $\text{Boy}_1, \text{Girl}, \text{Girl}, \text{Girl}, \text{Boy}_2$.
2. There are $2$ boys to place in the $2$ end seats:
   - There are $2$ choices for the leftmost seat.
   - There is $1$ remaining choice for the rightmost seat.
   - Number of ways to seat the boys: $2! = 2 \times 1 = 2$.
3. There are $3$ girls to place in the $3$ middle seats:
   - Number of ways to seat the girls: $3! = 3 \times 2 \times 1 = 6$.
4. By the Fundamental Counting Principle, the total number of arrangements is:
   $$2 \times 6 = 12$$

ANSWER 3: E

---

### Problem 4:
**Restatement:** 
Abe has $1$ green and $1$ red jelly bean ($2$ total). Bob has $1$ green, $1$ yellow, and $2$ red jelly beans ($4$ total). Each randomly picks one jelly bean. What is the probability that the colors match?

**Step-by-step solution:**
1. The matching colors can only be green or red (since Abe has no yellow jelly beans).
2. **Case 1: Both pick green:**
   $$P(\text{Abe picks green}) = \frac{1}{2}$$
   $$P(\text{Bob picks green}) = \frac{1}{4}$$
   $$P(\text{Both green}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$$
3. **Case 2: Both pick red:**
   $$P(\text{Abe picks red}) = \frac{1}{2}$$
   $$P(\text{Bob picks red}) = \frac{2}{4} = \frac{1}{2}$$
   $$P(\text{Both red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{2}{8}$$
4. **Total probability:**
   $$P(\text{Match}) = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** 
We need to find the product of the digits of the first palindrome year strictly after $2002$.

**Step-by-step solution:**
1. A four-digit palindrome has the form $abba$.
2. For the year to be after $2002$ and within the 2000s, the first digit $a$ must be $2$, which means the last digit $a$ is also $2$.
3. The year is of the form $2bb2$:
   - For $b = 0$, the year is $2002$.
   - The next smallest choice for $b$ greater than $0$ is $b = 1$, giving the year $2112$.
4. The product of the digits of $2112$ is:
   $$2 \times 1 \times 1 \times 2 = 4$$

ANSWER 5: B

---

### Problem 6:
**Restatement:** 
On a map, $12\text{ cm}$ represents $72\text{ km}$. We need to find how many kilometers a $17\text{ cm}$ length represents.

**Step-by-step solution:**
1. Find the scale factor in kilometers per centimeter:
   $$\text{Scale} = \frac{72\text{ km}}{12\text{ cm}} = 6\text{ km/cm}$$
2. Multiply this rate by the given length of $17\text{ cm}$:
   $$\text{Distance} = 17\text{ cm} \times 6\text{ km/cm} = 102\text{ km}$$

ANSWER 6: B

---

### Problem 7:
**Restatement:** 
Paul pays a debt of $35$ cents using $5$-cent (nickels), $10$-cent (dimes), and/or $25$-cent (quarters) coins. We want to find the difference between the maximum and minimum number of coins he can use.

**Step-by-step solution:**
1. **Largest number of coins:**
   - To maximize the number of coins, Paul should use the coin with the smallest value, which is $5$ cents:
   $$35 \div 5 = 7\text{ coins (all nickels)}$$
2. **Smallest number of coins:**
   - To minimize the number of coins, Paul should use as many high-value coins as possible.
   - He cannot pay $35$ cents with $1$ coin (since the maximum denomination is $25$ cents).
   - Can he use $2$ coins? Yes: $1\text{ quarter } (25¢) + 1\text{ dime } (10¢) = 35¢$.
   - Thus, the minimum number of coins is $2$.
3. **Difference:**
   $$\text{Difference} = 7 - 2 = 5$$

ANSWER 7: E

---

### Problem 8:
**Restatement:** 
The mean, median, and unique mode of the seven positive integers $3, 4, 5, 6, 6, 7,$ and $x$ are all equal. We need to determine the value of $x$.

**Step-by-step solution:**
1. In the list $\{3, 4, 5, 6, 6, 7\}$, the number $6$ already appears twice, while all other numbers appear once.
2. If $x \neq 6$, then either $x$ matches another number (e.g., if $x=5$, both $5$ and $6$ appear twice, so the mode would not be unique) or all other numbers appear once, leaving $6$ as the unique mode. If $x = 6$, $6$ appears three times, making $6$ the unique mode. In all valid cases where a unique mode exists, that unique mode must be $6$.
3. Since the mean, median, and unique mode are all equal, the mean must also be $6$.
4. Using the definition of the mean for the $7$ numb

## Problem 1:
The average of the four scores is 70, so the total of all four scores is 4 × 70 = 280.

The three known scores sum to 70 + 80 + 90 = 240.

The remaining score is 280 − 240 = 40.

Check: 40, 70, 80, 90 average to 280/4 = 70. ✓

ANSWER 1: A

## Problem 2:
Let the total bill be B. With 8 friends splitting equally, each should pay B/8.

Since Judi paid nothing, the other 7 friends each paid B/7 instead. The extra each paid is:

B/7 − B/8 = B/56

We're told this extra is $2.50, so B/56 = 2.50, giving B = 56 × 2.50 = $140.

Check: Each of the 7 pays $140/7 = $20, which is $2.50 more than the fair share of $140/8 = $17.50. ✓

ANSWER 2: C

## Problem 3:
A boy must sit at each end and the three girls in the middle.

- The two boys can swap between the two end seats: 2! = 2 ways.
- The three girls can be arranged in the three middle seats: 3! = 6 ways.

Total arrangements: 2 × 6 = 12.

ANSWER 3: E

## Problem 4:
Abe has 1 green and 1 red, so P(Abe green) = 1/2 and P(Abe red) = 1/2.
Bob has 1 green, 1 yellow, 2 red (4 total), so P(Bob green) = 1/4, P(Bob red) = 2/4 = 1/2.

The colors match only if both show green or both show red:

P(both green) = (1/2)(1/4) = 1/8
P(both red) = (1/2)(1/2) = 1/4

Total: 1/8 + 2/8 = 3/8.

ANSWER 4: C

## Problem 5:
We need the next palindrome year after 2002. Years of the form ABBA between 2002 and 3000 start with 2, so they have the form 2BB2.

The next after 2002 is 2112 (with 2112 > 2002, and 2 0 0 2 already used).

Product of digits: 2 × 1 × 1 × 2 = 4.

ANSWER 5: B

## Problem 6:
The map scale: 12 cm represents 72 km, so each centimeter represents 72 ÷ 12 = 6 km.

A 17 cm length represents 17 × 6 = 102 km.

ANSWER 6: B

## Problem 7:
Paul must pay exactly 35 cents using 5-, 10-, and 25-cent coins.

**Largest number of coins:** Use the smallest coins: seven 5-cent coins = 7 coins.

**Smallest number of coins:** 25 + 10 = 35, which uses just 2 coins. (One coin can't work since no coin is worth 35 cents.)

Difference: 7 − 2 = 5.

ANSWER 7: E

## Problem 8:
The mode must be 6 (already appearing twice), and it must remain unique.

Test x = 11: the list is 3, 4, 5, 6, 6, 7, 11.
- Mean: (3+4+5+6+6+7+11)/7 = 42/7 = 6 ✓
- Median (middle of 7 sorted values): 6 ✓
- Mode: 6 appears twice, all others once ✓

All three measures equal 6, so x = 11.

(Note: x = 6 gives mean 37/7 ≠ 6; x = 5 makes 5 appear twice, breaking the unique mode.)

ANSWER 8: D

## Problem 9:
Susan spent $12 on food and twice as much on rides: 2 × 12 = $24.

Total spent: 12 + 24 = $36.

Money left: 50 − 36 = $14.

ANSWER 9: B

## Problem 10:
Following order of operations (division before addition and subtraction):

16 + 8 ÷ 4 − 2 = 16 + 2 − 2 = 16.

(The tempting wrong answer 4 comes from computing left to right: (16+8) ÷ 4 − 2 = 4, which ignores order of operations.)

ANSWER 10: D

Problem 1:
We need the fourth exam score. The average of four scores is 70, so their total must be
\[
4\cdot 70=280.
\]
The three known scores sum to
\[
70+80+90=240.
\]
Therefore the remaining score is
\[
280-240=40.
\]
A tempting wrong choice might be 70, but that would make the average 80.

ANSWER 1: A

Problem 2:
Let the total bill be \(T\). Since 8 friends were supposed to split equally, Judi’s share would have been
\[
\frac{T}{8}.
\]
Instead, her 7 friends each paid an extra \(\$2.50\), so together they paid Judi’s share:
\[
7(2.50)=17.50.
\]
Thus
\[
\frac{T}{8}=17.50,
\]
so
\[
T=8(17.50)=140.
\]
A common mistake is to multiply only the extra amounts by 8, but the extra money equals Judi’s one share, not the whole bill.

ANSWER 2: C

Problem 3:
The team has 2 boys and 3 girls. The row must have a boy at each end and the 3 girls in the middle.

- The 2 boys can be placed at the two ends in
\[
2!=2
\]
ways.
- The 3 girls can be arranged in the middle in
\[
3!=6
\]
ways.

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

ANSWER 3: E

Problem 4:
Abe has 1 green and 1 red jelly bean, so he is equally likely to show green or red.

Bob has 1 green, 1 yellow, and 2 red jelly beans, so there are 4 equally likely jelly beans he could show if we distinguish the two reds.

The colors match in these cases:
- Abe shows green and Bob shows green:
\[
\frac12\cdot\frac14=\frac18
\]
- Abe shows red and Bob shows red:
\[
\frac12\cdot\frac24=\frac14
\]

Total probability:
\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

ANSWER 4: C

Problem 5:
The next palindrome after 2002 is 2112. To see this, 4-digit palindromes have the form \(abba\). After 2002, the next one with the same first digit 2 is 2112.

The product of its digits is
\[
2\cdot 1\cdot 1\cdot 2=4.
\]
A tempting wrong answer is 0, which is the product of the digits of 2002, not the next palindrome.

ANSWER 5: B

Problem 6:
On the map, 12 centimeters represents 72 kilometers, so 1 centimeter represents
\[
\frac{72}{12}=6
\]
kilometers.

Therefore 17 centimeters represents
\[
17\cdot 6=102
\]
kilometers.

ANSWER 6: B

Problem 7:
Paul must pay 35 cents.

Smallest number of coins: use a 25-cent coin and a 10-cent coin:
\[
25+10=35,
\]
so the minimum is 2 coins.

Largest number of coins: use only 5-cent coins:
\[
35\div 5=7,
\]
so the maximum is 7 coins.

The difference is
\[
7-2=5.
\]

ANSWER 7: E

Problem 8:
The numbers are
\[
3,4,5,6,6,7,x.
\]
The mode must be unique and equal to the mean and median.

The number 6 already appears twice. If \(x\) is 3, 4, 5, or 7, then two numbers would appear twice, so there would not be a unique mode. If \(x=6\), then the mean is
\[
\frac{3+4+5+6+6+6+7}{7}=\frac{37}{7},
\]
which is not 6. So the unique mode must be 6, and the common value of mean, median, and mode must be 6.

The sum of the known numbers is
\[
3+4+5+6+6+7=31.
\]
For the mean to be 6,
\[
\frac{31+x}{7}=6.
\]
Thus
\[
31+x=42,
\]
so
\[
x=11.
\]
Check: the sorted list is \(3,4,5,6,6,7,11\), whose median is 6, unique mode is 6, and mean is 6.

ANSWER 8: D

Problem 9:
Susan starts with 50 dollars.

She spends 12 dollars on food. She spends twice as much on rides:
\[
2\cdot 12=24.
\]
Total spent:
\[
12+24=36.
\]
Amount left:
\[
50-36=14.
\]

ANSWER 9: B

Problem 10:
Interpreting the missing symbol as division, the expression is
\[
16+8\div 4-2.
\]
Using order of operations, divide first:
\[
8\div 4=2.
\]
Then compute:
\[
16+2-2=16.
\]
A tempting wrong answer is 4 if one incorrectly groups the addition before dividing:
\[
(16+8)\div 4-2=4.
\]

ANSWER 10: D

# Problem 1:
**Restate:** Four scores; three are 70, 80, 90; the average of all four is 70. Find the fourth.

**Set up:** Average 70 for 4 scores means total = 4 × 70 = 280.

Sum of known scores = 70 + 80 + 90 = 240.

Missing score = 280 − 240 = **40**.

*Ruling out E (70):* if the fourth were 70, the total would be 310 and the average 77.5, not 70. The three known scores average 80, which is above 70, so the fourth must be well below 70.

ANSWER 1: A

# Problem 2:
**Restate:** 8 people split a bill equally; one (Judi) pays nothing, so the other 7 each add $2.50 to cover her share. Find the total bill.

**Set up:** Let the total bill be $B$. Each person's fair share is $B/8$ — that's exactly the amount Judi didn't pay.

The 7 friends covered that share: $7 \times \$2.50 = \$17.50 = B/8$.

So $B = 8 \times 17.50 = \$140$.

*Check:* $140/8 = 17.50$ per person; each of 7 friends paid $17.50 + 2.50 = \$20$, and $7 \times 20 = \$140$. ✓

*Ruling out A ($120):* that would make each share $15, needing $15/7 ≈ \$2.14$ extra, not $2.50.

ANSWER 2: C

# Problem 3:
**Restate:** 2 boys, 3 girls in a row; boys on the two ends, girls in the three middle seats. Count arrangements.

**Set up:** Ends: choose which boy is on the left — 2 ways (the other boy takes the right end), i.e. $2! = 2$.

Middle three seats: arrange 3 girls in $3! = 6$ ways.

Total $= 2 \times 6 = 12$.

*Ruling out A (2) and D (6):* 2 counts only the boys' swap, 6 counts only the girls' arrangements; we must multiply both.

ANSWER 3: E

# Problem 4:
**Restate:** Abe has {G, R}; Bob has {G, Y, R, R}. Each shows one random bean. Probability the colors match.

**Set up:** Abe: G with prob 1/2, R with prob 1/2. Bob: G with prob 1/4, Y with 1/4, R with 2/4 = 1/2.

Matching cases:
- Both green: $\tfrac12 \cdot \tfrac14 = \tfrac18$
- Both red: $\tfrac12 \cdot \tfrac12 = \tfrac14$
- Yellow can never match (Abe has no yellow).

Total $= \tfrac18 + \tfrac14 = \tfrac18 + \tfrac28 = \tfrac38$.

*Ruling out D (1/2):* that would ignore that Bob's yellow (and the extra red weighting) changes his color probabilities from 1/2–1/2.

ANSWER 4: C

# Problem 5:
**Restate:** Find the first palindrome year after 2002, then multiply its digits.

**Set up:** A 4-digit palindrome looks like $\overline{abba}$. After 2002 (a = 2, b = 0), the next choice is $b = 1$: 2112. (Years 2003–2111 have no palindrome, since with a = 2 the last digit must be 2 and the third digit must equal the second: 2002, 2112, 2222, …)

Product of digits of 2112: $2 \times 1 \times 1 \times 2 = 4$.

*Ruling out A (0):* 0 would come from 2002 itself (or 2020, which isn't a palindrome).

ANSWER 5: B

# Problem 6:
**Restate:** 12 cm ↔ 72 km. How many km does 17 cm represent?

**Set up (unit rate):** $72 \div 12 = 6$ km per cm.

Then $17 \times 6 = 102$ km.

*Proportion check:* $\frac{12}{72} = \frac{17}{x} \Rightarrow 12x = 1224 \Rightarrow x = 102$. ✓

*Ruling out E (1224):* that's the cross-product before dividing by 12. *Ruling out A (6):* that's just the rate per cm.

ANSWER 6: B

# Problem 7:
**Restate:** Make 35 cents using 5s, 10s, 25s. Find (max # coins) − (min # coins).

**Most coins:** use all 5-cent coins: $35 \div 5 = 7$ coins.

**Fewest coins:** 25 + 10 = 35, so 2 coins. (One coin is impossible; no 35-cent coin.)

Difference $= 7 - 2 = 5$.

*Ruling out D (4):* that would come from thinking the minimum is 3 coins (e.g., 25+5+5), but 25+10 uses only 2.

ANSWER 7: E

# Problem 8:
**Restate:** For 3, 4, 5, 6, 6, 7, x (7 positive integers), mean = median = unique mode. Find x.

**Mode:** For the mode to be unique, 6 must be it (6 already appears twice; if x equaled another value like 5, we'd have two modes 5 and 6 — not unique; x = 6 would make mode 6 but let's test the mean).

So mode = 6, hence mean = median = 6.

**Mean condition:** $3+4+5+6+6+7 = 31$, so $\frac{31+x}{7} = 6 \Rightarrow 31 + x = 42 \Rightarrow x = 11$.

**Check median:** sorted: 3, 4, 5, **6**, 6, 7, 11 → median = 6 ✓; mode = 6 (unique) ✓; mean = 42/7 = 6 ✓.

*Ruling out B (6):* then the sum is 37, mean $37/7 \ne 6$. *Ruling out A (5):* mode would tie between 5 and 6.

ANSWER 8: D

# Problem 9:
**Restate:** Susan starts with $50, spends $12 on food and twice that on rides.

Rides: $2 \times 12 = \$24$.

Total spent: $12 + 24 = \$36$.

Left: $50 - 36 = \$14$.

*Ruling out C (26):* that's $50 − 24$, forgetting the food; *A (12)* comes from wrongly using $12 for rides too ($50 − 12 − 12 − ...$ mis-computation).

ANSWER 9: B

# Problem 10:
**Restate:** Compute $\dfrac{16+8}{4-2}$.

**Numerator:** $16 + 8 = 24$.
**Denominator:** $4 - 2 = 2$.

$\dfrac{24}{2} = 12$.

*Ruling out the traps:* dividing only the 8 by 4 gives $16 + 2 - 2 = 16$ (D); dividing only by the whole $4-2$ gives $16 + 4 = 20$ (E); dividing the numerator by just 4 then subtracting 2 gives $6 - 2 = 4$ (A). The fraction bar groups the entire top and entire bottom, so the answer is 12.

ANSWER 10: C

Problem 1:

We are asked for the fourth exam score.

If the average of 4 scores is 70, then the total of all 4 scores is

\[
4\cdot 70=280.
\]

The three known scores total

\[
70+80+90=240.
\]

So the remaining score is

\[
280-240=40.
\]

The tempting choice \(60\) would make the total \(300\), whose average is \(75\), not \(70\).

ANSWER 1: A

Problem 2:

We are asked for the total restaurant bill.

Judi's 7 friends each paid an extra \$2.50 because Judi did not pay.

Thus, Judi's share of the bill was

\[
7\cdot \$2.50=\$17.50.
\]

Since 8 friends were supposed to split the bill equally, each person's share was \$17.50. Therefore, the total bill was

\[
8\cdot \$17.50=\$140.
\]

Choice B, \$128, would mean each person's share was only \$16, so the seven friends would need to pay an extra \(\$16/7\), not \$2.50.

ANSWER 2: C

Problem 3:

We need count arrangements with the 2 boys at the two ends and the 3 girls in the middle.

The two boys can be arranged at the ends in

\[
2!=2
\]

ways.

The three girls can be arranged in the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

\[
2!\cdot 3!=2\cdot 6=12.
\]

Choice D, \(6\), counts only the arrangements of the girls and forgets that the boys can switch ends.

ANSWER 3: E

Problem 4:

Abe has 1 green and 1 red jelly bean, so

\[
P(\text{Abe chooses green})=\frac12,\qquad
P(\text{Abe chooses red})=\frac12.
\]

Bob has 1 green, 1 yellow, and 2 red jelly beans, for 4 total:

\[
P(\text{Bob chooses green})=\frac14,\qquad
P(\text{Bob chooses red})=\frac24=\frac12.
\]

Their colors match in two possible ways:

- Both choose green:

\[
\frac12\cdot\frac14=\frac18
\]

- Both choose red:

\[
\frac12\cdot\frac12=\frac14
\]

So the total probability is

\[
\frac18+\frac14=\frac18+\frac28=\frac38.
\]

Yellow cannot produce a match because Abe has no yellow jelly bean.

ANSWER 4: C

Problem 5:

We seek the next palindrome after \(2002\).

A four-digit palindrome has the form

\[
abba.
\]

After \(2002\), the next palindrome is

\[
2112.
\]

The product of its digits is

\[
2\cdot 1\cdot 1\cdot 2=4.
\]

Choice A, \(0\), would be the product for \(2002\), but the problem asks for the next palindrome after \(2002\).

ANSWER 5: B

Problem 6:

A 12-centimeter map length represents 72 kilometers.

So 1 centimeter represents

\[
\frac{72}{12}=6
\]

kilometers.

Therefore, 17 centimeters represents

\[
17\cdot 6=102
\]

kilometers.

Choices such as \(204\) or \(864\) result from multiplying when we should first find the number of kilometers per centimeter.

ANSWER 6: B

Problem 7:

Paul must pay 35 cents using 5-cent, 10-cent, and 25-cent coins.

To use the largest possible number of coins, he should use all 5-cent coins:

\[
35\div 5=7
\]

coins.

To use the smallest possible number of coins, he can use

\[
25+10=35,
\]

which uses 2 coins.

The difference is

\[
7-2=5.
\]

A payment such as \(25+5+5\) uses 3 coins, but it is not the smallest possible number because \(25+10\) uses only 2.

ANSWER 7: E

Problem 8:

The numbers are

\[
3,4,5,6,6,7,x.
\]

The current unique mode is \(6\), since 6 appears twice and every other listed number appears once. Therefore, the common mean, median, and mode must all be 6.

For the mean to be 6, the total of all 7 numbers must be

\[
7\cdot 6=42.
\]

The known numbers add to

\[
3+4+5+6+6+7=31.
\]

Thus,

\[
31+x=42,
\]

so

\[
x=11.
\]

Checking: the ordered list is

\[
3,4,5,6,6,7,11.
\]

The median is 6, and the unique mode is 6, as required.

If \(x=6\), then the mode and median would be 6, but the mean would not be 6.

ANSWER 8: D

Problem 9:

Susan starts with \$50.

She spends \$12 on food. She spends twice that amount on rides:

\[
2\cdot 12=24.
\]

Her total spending is

\[
12+24=36.
\]

Therefore, she has left

\[
50-36=14
\]

dollars.

Choice C, \$26, is what she would have after buying food only, before paying for rides.

ANSWER 9: B

Problem 10:

The expression is

\[
16+\frac{8}{4}-2.
\]

Use order of operations: perform the division first.

\[
\frac84=2.
\]

Then compute

\[
16+2-2=16.
\]

A tempting error is to add \(16+8\) first, but division must be done before addition and subtraction.

ANSWER 10: D