🔬 Model Lab

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.

Problem 1:
We need the number of males who are in band but not orchestra.

Band total = 100 female + 80 male = 180.
Orchestra total = 80 female + 100 male = 180.

Let the overlap, students in both, have $F$ females and $M$ males.
Given $F=60$.

By inclusion-exclusion for union:
|Band $\cup$ Orchestra| = |Band| + |Orchestra| - |Both|
$230 = 180 + 180 - (F+M)$
$230 = 360 - (60+M)$
$60+M = 360-230 =130$
$M = 70$.

So 70 males are in both.

Males in band total 80. Those NOT in orchestra = total band males - both:
$80-70 = 10$.

This rules out B-E, which come from confusing females with males or forgetting to subtract.

ANSWER 1: A

Problem 2:
We need bags of rice equivalent to 1 fish.

Given: 3 fish = 2 loaves.
Given: 1 loaf = 4 bags.

So 2 loaves = $2 \times 4 = 8$ bags.

Hence: 3 fish = 8 bags.
Divide by 3: 1 fish = $8/3$ bags $= 2\frac{2}{3}$ bags.

Choices B, C are $<1$, impossible since a fish buys more than a loaf which buys 4 bags. A, E are miscalculations.

ANSWER 2: D

Problem 3:
We have 7 numbers $a_1,\dots,a_7$. First four average 5, last four average 8, all seven average $6\frac47$.

Sum first four $=4\times 5 =20$.
Sum last four $=4\times 8 =32$.
Sum all seven $=7\times 6\frac47 =7\times \frac{46}{7}=46$.

Let the middle number $a_4=m$ be the one common to both sets of four. Then adding the two 4-sums counts $m$ twice:
$20+32 = (\text{sum of all 7}) + m$
$52 = 46+m$
$m = 6$.

So the common number is 6, not the overall average, ruling out C, etc.

ANSWER 3: B

Problem 4:
9 coins, at least one penny(1), nickel(5), dime(10), quarter(25), total 102 cents. Find number of dimes.

Let counts be $p,n,d,q\ge 1$ with $p+n+d+q=9$ and $p+5n+10d+25q=102$.

Take out one of each: that uses 4 coins worth $1+5+10+25=41$.
Remaining: 5 coins must be worth $102-41=61$.

Let $p'=p-1$ etc, so $p'+n'+d'+q'=5$ and $p'+5n'+10d'+25q'=61$.

Subtract: $4n'+9d'+24q'=56$.

$q'$ can be 0,1,2.

- $q'=2$: $4n'+9d'=8$. Only $d'=0,n'=2$ works. Then $p'=1$. This gives $p=2,n=3,d=1,q=3$. Check: $2+3+1+3=9$ coins, $2+15+10+75=102$. Valid.
- $q'=1$: $4n'+9d'=32$. Try $d'=0,1,2,3$: gives $n'=8,23/4,7/2,5/4$, none valid with sum $\le5$.
- $q'=0$: $4n'+9d'=56$. Mod 4 forces $d'=0$ or $4$, giving $n'=14$ or $5$, both make $n'+d'+q'>5$, impossible.

Thus the only solution has $d=1$.

So more dimes would overshoot the value or coin count.

ANSWER 4: A

Problem 5:
24 numbers use 2,4,5,7 once each. One is a multiple of another. Identify the multiple (the larger).

Let small $\times k =$ large, both 4-digit. Smallest possible $\approx2457$, largest $\approx7542$, so ratio $k<4$. So $k=2$ or $3$.

If $k=2$, large must be even. If large is odd, $k$ must be 3.

Check each choice as potential large (it cannot be small if doubling already exceeds 9999):

A. 5724: $5724/2=2862$ no, $5724/3=1908$ no. As small, $5724\times2>9999$. Out.
B. 7245 odd: $7245/3=2415$ contains 1. Out. Cannot be $\times2$.
C. 7254: $7254/2=3627$ no, $7254/3=2418$ no. As small, $\times2>9999$. Out.
D. 7425 odd: $7425/3=2475$. 2475 uses digits 2,4,7,5 exactly! So $2475\times3=7425$. This works.
E. 7542: $7542/2=3771$ no, $7542/3=2514$ contains 1. Out.

Only D participates in such a pair.

ANSWER 5: D

Problem 6:
1200 students, each takes 5 classes. Each class has 30 students. How many class meetings? How many teachers if each teaches 4?

Total student-seats $=1200\times5=6000$.
Each class holds 30 students, so number of classes $=6000/30=200$.

Each teacher teaches 4 classes, so teachers $=200/4=50$.

30 would be ignoring the 5 vs 4 difference, 40 would be dividing wrong way, etc.

ANSWER 6: E

Problem 7:
Start 128. Sell 25% to Jill, then 25% of remainder to June, then give away 1. How many left?

First sale: $25\%$ of 128 $=32$. Remaining $=128-32=96$.
Second sale: $25\%$ of 96 $=24$. Remaining $=96-24=72$.
Give 1 to teacher: $72-1=71$.

Forgetting the final -1 gives 72; miscomputing percentages gives other choices.

ANSWER 7: D

Problem 8:
Five different positive integers, mean 15, median 18. Maximize the largest.

Sorted $a<b<c<d<e$, distinct.
Mean 15 $\implies$ sum $=5\times15=75$.
Median $c=18$. So $a+b+18+d+e=75$, i.e. $a+b+d+e=57$.

To make $e$ as large as possible, make $a,b,d$ as small as possible while respecting $a<b<18<d<e$ and distinct positive.

Smallest $a=1$, $b=2$. Smallest $d>18$ is $19$.
Then $e=57-(1+2+19)=57-22=35$.
Set $1,2,18,19,35$ works: different, median 18, mean 15.

Any larger $e$ would require smaller $a+b+d$, impossible.

ANSWER 8: D

Problem 9:
Total $T$. Alexa $T/4$, Brittany $2T/7$, Chelsea 15, others $O\le14$. Find $O$.

$T$ must be divisible by 4 and 7, so by 28. Let $T=28k$.

Remainder after Alexa and Brittany:
$T-T/4-2T/7 = T(1-1/4-2/7)=T(13/28)=13k$.

This equals $15+O$. So $13k=15+O$, $0\le O\le14$.

$k=1$: $13=15+O$ impossible.
$k=2$: $26=15+O\implies O=11$, which is $\le14$. Valid.
$k\ge3$: $13k-15\ge24>14$. Invalid.

Thus $O=11$.

ANSWER 9: B

Problem 10:
Digits 1-5 once each form PQRST with PQR divisible by 4, QRS divisible by 5, RST divisible by 3. Find P.

QRS divisible by 5 means its units digit $S$ is 0 or 5. Digits are 1-5, so $S=5$.

PQR divisible by 4 means two-digit QR divisible by 4. With digits from $\{1,2,3,4\}$, distinct, possibilities:
12 yes, 24 yes, 32 yes; 14,13,21,23,31,34,41,42,43 all no. So QR $\in\{12,24,32\}$.

Test each with $R+5+T$ divisible by 3:

- QR=12: Q=1,R=2. P,T are 3,4. Need $2+5+T=7+T\equiv0\pmod3\implies T\equiv2\pmod3$, but 3,4 are $0,1\pmod3$. Impossible.
- QR=24: Q=2,R=4. P,T are 1,3. Need $4+5+T=9+T\equiv0\pmod3\implies T\equiv0\pmod3$, so $T=3$, $P=1$. Check: 124 divisible by 4, 245 divisible by 5, 453 sum 12 divisible by 3. Works.
- QR=32: Q=3,R=2. P,T are 1,4. Need $7+T\equiv0\pmod3$, requires $T\equiv2\pmod3$, but 1,4 are $1\pmod3$. Impossible.

Thus unique solution $P=1$.

ANSWER 10: A

An expert step-by-step solution for each problem is presented below.

---

### Problem 1:
**Restatement:**
We are given:
- Band: 100 females, 80 males (Total = 180)
- Orchestra: 80 females, 100 males (Total = 180)
- Students in both: 60 females
- Total unique students in either or both: 230

We want to find the number of males in the band who are *not* in the orchestra.

**Step-by-step Solution:**
1. Using the Principle of Inclusion-Exclusion for the total number of students:
   $$\text{Total} = \text{Band} + \text{Orchestra} - \text{Both}$$
   $$230 = 180 + 180 - \text{Both} \implies \text{Both} = 360 - 230 = 130$$
2. The number of students in both consists of both females and both males:
   $$\text{Both Males} = \text{Both} - \text{Both Females} = 130 - 60 = 70$$
3. The number of males in the band who are NOT in the orchestra is:
   $$\text{Band Males} - \text{Both Males} = 80 - 70 = 10$$

ANSWER 1: A

---

### Problem 2:
**Restatement:**
We are given two trade ratios:
- $3\text{ fish} = 2\text{ loaves of bread}$
- $1\text{ loaf of bread} = 4\text{ bags of rice}$

We need to determine how many bags of rice one fish is worth.

**Step-by-step Solution:**
1. From the second relation, 2 loaves of bread equal:
   $$2 \times 4\text{ bags of rice} = 8\text{ bags of rice}$$
2. Substituting into the first relation:
   $$3\text{ fish} = 8\text{ bags of rice}$$
3. Dividing both sides by 3 to find the value of 1 fish:
   $$1\text{ fish} = \frac{8}{3} = 2\frac{2}{3}\text{ bags of rice}$$

ANSWER 2: D

---

### Problem 3:
**Restatement:**
There are seven numbers $x_1, x_2, \dots, x_7$:
- The average of the first four ($x_1, x_2, x_3, x_4$) is $5$.
- The average of the last four ($x_4, x_5, x_6, x_7$) is $8$.
- The average of all seven is $6\frac{4}{7}$.

We want to find the number common to both sets, which is $x_4$.

**Step-by-step Solution:**
1. Calculate the sum of the first four numbers:
   $$\text{Sum}_1 = 4 \times 5 = 20$$
2. Calculate the sum of the last four numbers:
   $$\text{Sum}_2 = 4 \times 8 = 32$$
3. Calculate the sum of all seven numbers:
   $$\text{Total Sum} = 7 \times 6\frac{4}{7} = 7 \times \frac{46}{7} = 46$$
4. Notice that adding $\text{Sum}_1$ and $\text{Sum}_2$ includes every number once, except $x_4$, which is included twice:
   $$\text{Sum}_1 + \text{Sum}_2 = \text{Total Sum} + x_4$$
   $$20 + 32 = 46 + x_4$$
   $$52 = 46 + x_4 \implies x_4 = 6$$

ANSWER 3: B

---

### Problem 4:
**Restatement:**
You have a total of 9 coins consisting of pennies ($p$), nickels ($n$), dimes ($d$), and quarters ($q$), with at least one of each coin. The total value is $\$1.02 = 102$ cents. Find the number of dimes.

**Step-by-step Solution:**
1. Set up the equations:
   $$p + n + d + q = 9$$
   $$p + 5n + 10d + 25q = 102$$
   where $p, n, d, q \ge 1$ are positive integers.
2. Consider the value modulo 5:
   $$p \equiv 102 \equiv 2 \pmod 5$$
   Since $p \ge 1$ and the sum of all four positive integers is 9, $p$ can only be $2$ (if $p = 7$, then $n+d+q = 2$, which contradicts each being at least 1).
3. Substitute $p = 2$ into the equations:
   $$n + d + q = 7$$
   $$5n + 10d + 25q = 100 \implies n + 2d + 5q = 20$$
4. Subtract the first simplified equation from the second:
   $$(n + 2d + 5q) - (n + d + q) = 20 - 7$$
   $$d + 4q = 13$$
5. Test possible values for $q \ge 1$:
   - If $q = 1$: $d = 9$, which exceeds $n+d+q=7$.
   - If $q = 2$: $d = 5$, then $n = 7 - 5 - 2 = 0$, but $n \ge 1$, so this is invalid.
   - If $q = 3$: $d = 13 - 12 = 1$, then $n = 7 - 1 - 3 = 3$. This works!
   - If $q \ge 4$: $4q \ge 16 > 13$, which gives a negative $d$.
6. Therefore, the only valid solution is $p = 2, n = 3, d = 1, q = 3$, meaning there is $1$ dime.

ANSWER 4: A

---

### Problem 5:
**Restatement:**
Among all 24 four-digit numbers using the digits 2, 4, 5, and 7, exactly one is a multiple of another. We need to identify which of the given choices is that multiple.

**Step-by-step Solution:**
1. The smallest possible number made from these digits is $2457$, and the largest is $7542$.
2. The ratio of the multiple to the original number must be an integer:
   - A ratio of 2: since the smallest number is $2457$, $2 \times 2457 = 4914$ (not using the digits).
   - A ratio of 3: testing numbers starting with 2:
     $$2475 \times 3 = 7425$$
3. Both $2475$ and $7425$ consist of the digits 2, 4, 5, and 7 each used exactly once.
4. Thus, $7425$ is a multiple of $2475$, and it is listed as choice D.

ANSWER 5: D

---

### Problem 6:
**Restatement:**
- 1200 students, each takes 5 classes a day.
- Each teacher teaches 4 classes.
- Each class has 30 students and 1 teacher.
Find the number of teachers.

**Step-by-step Solution:**
1. Find the total number of student-class seats needed each day:
   $$\text{Total student-classes} = 1200 \times 5 = 6000$$
2. Since each class has 30 students, the number of class sections needed is:
   $$\text{Class sections} = \frac{6000}{30} = 200$$
3. Each class section requires 1 teacher, so 200 teaching periods must be covered.
4. Each teacher teaches 4 classes per day, so the number of teachers needed is:
   $$\text{Teachers} = \frac{200}{4} = 50$$

ANSWER 6: E

---

### Problem 7:
**Restatement:**
Jack starts with 128 apples:
1. Sells 25% to Jill.
2. Sells 25% of the remainder to June.
3. Gives 1 apple to his teacher.
Find how many apples he has left.

**Step-by-step Solution:**
1. Selling 25% leaves $75\% = \frac{3}{4}$ of the apples:
   $$\text{Remaining after Jill} = 128 \times \frac{3}{4} = 96$$
2. Selling 25% of the remainder leaves $\frac{3}{4}$ again:
   $$\text{Remaining after June} = 96 \times \frac{3}{4} = 72$$
3. Giving 1 apple to the teacher leaves:
   $$72 - 1 = 71$$

ANSWER 7: D

---

### Problem 8:
**Restatement:**
A set of five distinct positive integers has a mean of 15 and a median of 18. Find the maximum possible value of the largest integer.

**Step-by-step Solution:**
1. Let the five distinct positive integers in increasing order be:
   $$a < b < c < d < e$$
2. T

Problem 1:
Band has 100 female + 80 male = 180 members; orchestra has 80 + 100 = 180 members. Using inclusion–exclusion on all 230 students:
180 + 180 − (both) = 230, so **both = 130**.
Of the 130 students in both, 60 are female, so 70 are males in both.
Males in band only = 80 − 70 = **10**.
(Not 20 or 50 — those would come from miscounting the overlap.)

ANSWER 1: A

Problem 2:
3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags.
Thus 3 fish = 8 bags of rice, so 1 fish = 8/3 bags ≈ 2⅔.
(3/4 would be dividing incorrectly; the chain is fish → loaves → rice.)

ANSWER 2: E

Problem 3:
Sum of first four numbers: 4 × 5 = 20.
Sum of last four numbers: 4 × 8 = 32.
Sum of all seven: 7 × 46/7 = 46.
The fourth number is counted in both groups of four, so 20 + 32 − x = 46, giving x = 6.

ANSWER 3: B

Problem 4:
Total value $1.02 ends in 2, and only pennies affect the last digit, so the number of pennies is 2, 7, 12, ...
Try 2 quarters: remaining $0.52 with 7 coins. Let d, n, p be dimes, nickels, pennies:
d + n + p = 7 and 10d + 5n + p = 52. Subtracting: 9d + 4n = 45. Solutions (d,n): (1,9), (5,0) — both impossible (too many coins or no nickel).
Try 3 quarters: remaining $0.27 with 6 coins: 10d + 5n + p = 27, d + n + p = 6 → 9d + 4n = 21. Try d=1, n=3: 9+12=21 ✓, p = 2. ✓ (one of each type, 1+3+2+3 = 9 coins, value 10+15+2+75 = 102 ✓).
So there must be exactly **1 dime**.

ANSWER 4: A

Problem 5:
A multiple must be 2, 3, 5, or 7 times another (since both are 4-digit; 2457×... ratios between numbers range roughly 7425/2457 ≈ 3). Test each choice:
- 5724 = 2 × 2862 (digits wrong); /3 = 1908; /7 = 817.7 — no.
- 7245 = 3 × 2415 (repeats 2,4) ; /5 = 1449 — no.
- 7254 = 2×3627, 3×2418 — no.
- 7425 = 3 × 2475, and 2475 uses digits 2,4,7,5 exactly once. ✓
- 7542 = 2×3771, 3×2514 (repeats? 2,5,1,4 — wrong digits) — no.
So 7425 = 3 × 2475.

ANSWER 5: D

Problem 6:
Total class-enrollments per day: 1200 students × 5 classes = 6000 student-seats.
Each class has 30 students, so number of classes = 6000/30 = 200.
Each teacher teaches 4 classes, so teachers = 200/4 = 50.
(Not 40 — that forgets each teacher teaches 4, not 5, classes.)

ANSWER 6: E

Problem 7:
Start: 128. Sell 25% (32) to Jill → 96 remain.
Sell 25% of 96 (24) to June → 72 remain.
Give 1 to teacher → 71.
(71, not 72 — don't forget the teacher's apple.)

ANSWER 7: D

Problem 8:
Sum of the five integers = 5 × 15 = 75. Median 18 means the middle (third) number is 18.
To maximize the largest, minimize the other four while keeping them distinct positive integers below/around 18: 1, 2, 18, and then the fourth must exceed 18, so its minimum is 19.
Then the largest = 75 − (1 + 2 + 18 + 19) = 75 − 40 = 35.

ANSWER 8: D

Problem 9:
The total T must be divisible by both 4 and 7, so T is a multiple of 28.
Others scored = T − (T/4) − (2T/7) − 15 = 13T/28 − 15, and this must be between 0 and 14.
- T = 28: 13 − 15 < 0. ✗
- T = 56: 26 − 15 = 11, and 11 ≤ 7×2 = 14. ✓
- T = 84: 39 − 15 = 24 > 14. ✗
So the other 7 members scored 11 points.

ANSWER 9: B

Problem 10:
- QRS divisible by 5 forces S = 5. Remaining digits: {1,2,3,4} for P, Q, R, T.
- PQR divisible by 4 requires the two-digit number QR to be divisible by 4, using digits from {1,2,3,4} with Q ≠ R. Options: QR = 12, 24, 32 (44 repeats a digit). So R = 2 or 4.
- RST divisible by 3: R + 5 + T must be divisible by 3.
  - If R = 2: need 7 + T divisible by 3 → T ∈ {2, 5}, both already used. Impossible.
  - If R = 4: need 9 + T divisible by 3 → T = 3.
Then QR = 24 (Q = 2), S = 5, T = 3, and the remaining digit P = 1.
Check: PQR = 124 (÷4 ✓), QRS = 245 (÷5 ✓), RST = 453 (4+5+3=12, ÷3 ✓).

ANSWER 10: A

Problem 1:
Let me set up a table. Band: 100 F + 80 M = 180 total. Orchestra: 80 F + 100 M = 180 total. Union = 230.
Intersection = 180 + 180 − 230 = 130 total students in both.
Since 60 females are in both, males in both = 130 − 60 = 70.
Band males in orchestra = 70, so band males NOT in orchestra = 80 − 70 = 10.
Check: females only-band = 40, only-orchestra = 20 → total F = 120; males = 10 + 30 + 70 = 110; grand total = 230 ✓.
ANSWER 1: A

Problem 2:
3 fish = 2 loaves. Since 1 loaf = 4 bags of rice, 2 loaves = 8 bags.
So 3 fish = 8 bags → 1 fish = 8/3 = 2⅔ bags of rice.
ANSWER 2: D

Problem 3:
Let the middle (common) number be x.
Sum of first four = 4·5 = 20. Sum of last four = 4·8 = 32.
Sum of all seven = 7·(6 + 4/7) = 42 + 4 = 46.
Adding the two sums counts the middle number twice, so 20 + 32 = 46 + x → x = 52 − 46 = 6.
ANSWER 3: B

Problem 4:
Let p, n, d, q ≥ 1 be pennies, nickels, dimes, quarters (values in cents).
p + n + d + q = 9 and p + 5n + 10d + 25q = 102.
Subtracting: 4n + 9d + 24q = 93.
- q = 4: 96 > 93, too big; q ≥ 4 impossible.
- q = 3: need 4n + 9d = 21 → d ≡ 1 (mod 4). d = 1 gives n = 3. Then p = 9 − 3 − 1 − 3 = 2 ≥ 1 ✓.
- q = 2: need 4n + 9d = 45 → d ≡ 1 (mod 4): d = 1 → n = 9 (total coins ≥ 12, too many); d = 5 → n = 0, invalid.
- q = 1: need 4n + 9d = 69 → d = 1 → n = 15; d = 5 → n = 6; both make p < 0.
Unique solution: p = 2, n = 3, d = 1, q = 3. Check: value = 2 + 15 + 10 + 75 = 102 ✓. So 1 dime.
ANSWER 4: A

Problem 5:
Every such number has digit sum 2 + 4 + 5 + 7 = 18, so all are multiples of 9; ratios of distinct numbers ≤ 7954/2345 ≈ 3.4, so a quotient can only be 2 or 3.
Numbers ending in 5 are multiples of 45: 45×{55, 61, 95, 105, 161, 165} = {2475, 2745, 4275, 4725, 7245, 7425}. Among the multipliers, only 165/55 = 3 works, giving 7425 = 3 × 2475.
Now test options: 5724/2 = 2862, 5724/3 = 1908 (bad digits); 7245/3 = 2415 (digit 1); 7254/2 =  Whether3627, /3 = 2418; 7542/2 = 3771, /3 = 2514 — none are in the set.
So the unique pair is (2475, 7425); the number is 7425.
ANSWER 5: D

Problem 6:
Total student-class assignments per day = 1200 × 5 = 6000.
Each class holds 30 students → classes per day = 6000/30 = 200.
Each teacher teaches 4 classes → teachers = 200/4 = 50.
ANSWER 6: E

Problem 7:
After Jill: 128 × (1 − 0.25) = 96. After June: 96 × 0.75 = 72. After giving away 1: 72 − 1 = 71.
ANSWER 7: D

Problem 8:
Sorted integers a < b < c < d < e, with c = 18 and sum = 5 × 15 = 75.
To maximize e, minimize a, b, d: smallest choices a = 1, b = 2, and smallest d > 18 is 19.
Then e = 75 − 1 − 2 − 18 − 19 = 35. (Set {1, 2, 18, 19, 35} works.)
ANSWER 8: E

Problem 9:
Let total = N. Alexa = N/4, Brittany = 2N/7, so 28 | N.
Others total = N − 7 − 15 = (28 − 7 − 8)N/28 − 15 = 13N/28 − 15, and each of the 7 ≤ 2 → sum ≤ 14 and ≥ 0.
N = 28: others = −2 ✗. N = 56: Alexa = 14, Brittany = 16, others = 56 − 14 − 16 − 15 = 11 ✓ (≤ 14). N = 84: others = 24 > 14 ✗.
ANSWER 9: B

Problem 10:
QRS divisible by 5 → S = 5. PQR divisible by 4 → QR divisible by 4, with digits from {1,2,3,4}: only QR ∈ {12, 24, 32}.
Condition R + S + T ≡ 0 (mod 3) means R + T ≡ 1 (mod 3).
- QR = 12: {P,T} = {3,4}; need T ≡ 2 (mod 3) — neither 3 nor 4 works.
- QR = 32: {P,T} = {1,4}; need T ≡ 2 (mod 3) — neither works.
- QR = 24: {P,T} = {1,3}; need T ≡ 0 (mod 3) → T = 3, so P = 1.
Check: 12435 → 124 ÷ 4 = 31 ✓, 245 ÷ 5 = 49 ✓, 453 ÷ 3 = 151 ✓. Hence P = 1.
ANSWER 10: A

# Problem 1:

**Asked:** How many males are in band but not orchestra?

Band total = 100 + 80 = 180. Orchestra total = 80 + 100 = 180.

By inclusion–exclusion:
$$|B| + |O| - |B\cap O| = |B\cup O| \Rightarrow 180 + 180 - |B\cap O| = 230$$
So $|B\cap O| = 130$ students are in both.

Of those 130, 60 are female, so $130 - 60 = 70$ are male (in both).

Males in band = 80, so males in band only = $80 - 70 = 10$.

(Choice E = 70 is the "males in both" number; C = 30 mixes up the wrong totals.)

ANSWER 1: A

# Problem 2:

**Asked:** Value of one fish in bags of rice.

3 fish = 2 loaves, and 1 loaf = 4 bags of rice, so 2 loaves = 8 bags of rice.

Thus 3 fish = 8 bags of rice, so 1 fish = $\frac{8}{3} = 2\frac23$ bags.

(Choice E, $3\frac13$, would come from dividing wrongly; C, $\frac34$, inverts the ratio.)

ANSWER 2: D

# Problem 3:

**Asked:** The number shared by both groups of four.

Sum of first four = $4 \times 5 = 20$.
Sum of last four = $4 \times 8 = 32$.
Sum of these two sums = 52 — this counts all seven numbers once, plus the overlapping (4th) number twice.

Sum of all seven = $7 \times 6\frac47 = 7 \times \frac{46}{7} = 46$.

Common number = $52 - 46 = 6$.

ANSWER 3: B

# Problem 4:

**Asked:** Number of dimes among 9 coins totaling 102¢ with at least one of each type.

Only pennies contribute non‑multiples of 5, so the number of pennies must end so the total ends in 2: pennies $= 2$ or $7$.

- If pennies = 7: only 2 coins left, but we need a nickel, dime, and quarter — impossible.
- So pennies = 2, leaving 7 coins worth 100¢:
$$5n+10d+25q=100,\quad n+d+q=7,\ n,d,q\ge 1$$

Divide first by 5: $n+2d+5q=20$. Subtract $n+d+q=7$: $d+4q=13$.
- $q=1 \Rightarrow d=9$ (too many coins)
- $q=2 \Rightarrow d=5, n=0$ (no nickel — invalid)
- $q=3 \Rightarrow d=1, n=3$: coins $2+3+1+3=9$, value $2+15+10+75=102$ ✓

So exactly 1 dime.

ANSWER 4: A

# Problem 5:

**Asked:** Which listed number is a multiple of another rearrangement of 2,4,5,7?

A multiple must be a small integer times a number with the same digits. Test each:
- $7425 = 3 \times 2475$, and 2475 uses digits 2, 4, 7, 5 exactly once ✓

Check others fail: $5724/2 = 2862$ (wrong digits), $7245/3=2415$ (has a 1), $7254/3 = 2418$ (has 1,8), $7542/2=3771$, $7542/3=2514$ (has 1). None work.

ANSWER 5: D

# Problem 6:

**Asked:** Number of teachers.

Total student–class slots: $1200 \times 5 = 6000$.
Each class holds 30 students, so number of classes: $6000 / 30 = 200$.
Each teacher teaches 4 classes: $200 / 4 = 50$ teachers.

(Choice C = 40 comes from forgetting a step; the chain must go students → class slots → classes → teachers.)

ANSWER 6: E

# Problem 7:

**Asked:** Apples left after two 25% sales and giving away one.

Sold 25% to Jill: left $128 \times 0.75 = 96$.
Sold 25% of those to June: left $96 \times 0.75 = 72$.
Gave 1 to teacher: $72 - 1 = 71$.

(Choice B = 63 comes from subtracting 25% twice of the original; A = 7 is the sold-count confusion.)

ANSWER 7: D

# Problem 8:

**Asked:** Largest possible maximum value.

Sum $= 5 \times 15 = 75$. With five different positive integers $a<b<c<d<e$ and median $c=18$.

To maximize $e$, minimize the others: $a=1$, $b=2$, $c=18$, $d=19$ (must exceed 18, all distinct).

$e = 75 - (1+2+18+19) = 75 - 40 = 35$.

ANSWER 8: D

# Problem 9:

**Asked:** Points scored by the other 7 players.

Let $T$ = total. Since $\frac T4$ and $\frac{2T}{7}$ must be whole numbers, $T$ is a multiple of 28.

Others' points $= T - \frac T4 - \frac{2T}{7} - 15 = \frac{28T-7T-8T}{28} - 15 = \frac{13T}{28} - 15$.

Constraint: 7 players each scoring at most 2 points ⟹ others $\le 14$ (and $\ge 0$).

- $T=28$: $13-15 = -2$ — impossible.
- $T=56$: $26-15 = 11$ ✓ (fits, e.g., five 2's and one 1)
- $T=84$: $39-15 = 24 > 14$ — impossible.

So the answer is 11.

ANSWER 9: B

# Problem 10:

**Asked:** Find $P$ in $PQRST$ using digits 1–5 once each.

- $QRS$ divisible by 5 ⟹ $S = 5$ (0 isn't available).
- $PQR$ divisible by 4 ⟹ $R$ even, so $R \in \{2,4\}$.
- $RST$ divisible by 3 ⟹ $R + 5 + T$ divisible by 3.

If $R=2$: need $7+T \equiv 0 \pmod 3$ with $T \in \{1,3,4\}$: gives 8, 10, 11 — none divisible by 3.

So $R = 4$: need $9+T$ divisible by 3 with $T \in \{1,2,3\}$: $T=3$ (since 12 works).

Remaining digits for $P,Q$: $\{1,2\}$. Divisibility by 4 depends on $QR$: $14$ is not divisible by 4, $24$ is. So $Q=2$, $P=1$.

Check: $12435$ — $124/4=31$ ✓, $245/5=49$ ✓, $435/3=145$ ✓.

ANSWER 10: A

Problem 1:

We want the number of males who are in the band but **not** in the orchestra.

First find how many females are in at least one group:

\[
100+80-60=120
\]

The subtraction of 60 avoids counting the females in both groups twice.

Since there are 230 students total in the band or orchestra or both, the number of males in at least one group is

\[
230-120=110.
\]

There are 80 males in band and 100 males in orchestra. Let \(x\) be the number of males in both.

\[
80+100-x=110
\]

\[
180-x=110
\]

\[
x=70.
\]

Therefore, males in the band but not orchestra:

\[
80-70=10.
\]

Choices such as 20 or 30 would result from failing to account correctly for the males counted in both groups.

ANSWER 1: A

Problem 2:

Three fish can be traded for two loaves of bread.

Each loaf can be traded for four bags of rice, so two loaves are worth

\[
2\cdot4=8
\]

bags of rice.

Thus,

\[
3\text{ fish}=8\text{ bags of rice}.
\]

Divide by 3:

\[
1\text{ fish}=\frac83=2\frac23
\]

bags of rice.

The choices \(\frac38,\frac12,\frac34,\) and \(3\frac13\) do not equal \(\frac83\).

ANSWER 2: D

Problem 3:

The first four numbers have average 5, so their sum is

\[
4\cdot5=20.
\]

The last four numbers have average 8, so their sum is

\[
4\cdot8=32.
\]

The average of all seven numbers is \(6\frac47=\frac{46}{7}\), so the sum of all seven numbers is

\[
7\cdot\frac{46}{7}=46.
\]

When we add the sums of the first four and last four numbers, the one number common to both groups is counted twice:

\[
20+32=46+\text{common number}.
\]

So the common number is

\[
52-46=6.
\]

Thus values such as \(6\frac47\) are averages, not the repeated number itself.

ANSWER 3: B

Problem 4:

Let \(p,n,d,q\) be the numbers of pennies, nickels, dimes, and quarters.

There are 9 coins:

\[
p+n+d+q=9.
\]

Their value is 102 cents:

\[
p+5n+10d+25q=102.
\]

Subtract the first equation from the second:

\[
4n+9d+24q=93.
\]

Since there is at least one quarter, try possible values of \(q\). Since \(24q\le93\), we have \(q=1,2,\) or \(3\).

If \(q=1\),

\[
4n+9d=69.
\]

This gives too many coins or fails the coin-count condition.

If \(q=2\),

\[
4n+9d=45.
\]

The possible solution \(d=5,n=0\) has no nickels, which is not allowed.

If \(q=3\),

\[
4n+9d=21.
\]

Since there must be at least one dime and one nickel, the solution is

\[
d=1,\qquad n=3.
\]

Then

\[
p=9-3-1-3=2.
\]

Check the value:

\[
2(1)+3(5)+1(10)+3(25)=2+15+10+75=102.
\]

So there must be exactly 1 dime.

ANSWER 4: A

Problem 5:

Every number made from 2, 4, 5, and 7 has digit sum

\[
2+4+5+7=18,
\]

so each is divisible by 9.

The largest possible number is 7542 and the smallest is 2457. Therefore, if one such number is a multiple of another, the multiplier can only be 2 or 3.

Check the listed possibilities. The key one is

\[
2475\cdot3=7425.
\]

Both 2475 and 7425 use the digits 2, 4, 5, and 7 exactly once.

The other listed numbers do not work:

\[
5724\div2=2862,\qquad 5724\div3=1908
\]

\[
7245\div3=2415
\]

\[
7254\div2=3627,\qquad 7254\div3=2418
\]

\[
7542\div2=3771,\qquad 7542\div3=2514.
\]

None of those quotients uses exactly the digits 2, 4, 5, and 7 once.

Therefore the desired number is 7425.

ANSWER 5: D

Problem 6:

There are 1200 students, and each takes 5 classes per day.

So the total number of student class-seats is

\[
1200\cdot5=6000.
\]

Each class has 30 students, so the number of classes is

\[
\frac{6000}{30}=200.
\]

Each teacher teaches 4 classes, so the number of teachers is

\[
\frac{200}{4}=50.
\]

Choices such as 40 or 45 would not provide enough teacher-taught classes to cover all 200 classes.

ANSWER 6: E

Problem 7:

Jack begins with 128 apples.

He sells 25%, so he keeps 75%:

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

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

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

He then gives one apple to his teacher:

\[
72-1=71.
\]

A tempting error is to subtract 25% of the original 128 a second time; the second 25% is taken from the 96 apples remaining.

ANSWER 7: D

Problem 8:

Let the five different positive integers, in increasing order, be

\[
a<b<18<d<e.
\]

The median is 18, so it is the middle number.

The mean is 15, so the total of all five numbers is

\[
5\cdot15=75.
\]

To make the largest number \(e\) as large as possible, make the other four numbers as small as possible.

The smallest possible distinct positive values are

\[
a=1,\qquad b=2,\qquad d=19.
\]

Therefore,

\[
1+2+18+19+e=75.
\]

\[
40+e=75.
\]

\[
e=35.
\]

Any larger choices for \(a\), \(b\), or \(d\) would make \(e\) smaller.

ANSWER 8: D

Problem 9:

Let the total team score be \(T\).

Alexa scored \(\frac14T\), and Brittany scored \(\frac27T\). Chelsea scored 15 points.

Thus the other 7 players scored

\[
T-\frac14T-\frac27T-15.
\]

Combine the fractions:

\[
1-\frac14-\frac27
=\frac{28}{28}-\frac{7}{28}-\frac{8}{28}
=\frac{13}{28}.
\]

So the other 7 players scored

\[
\frac{13}{28}T-15.
\]

Because both \(\frac14T\) and \(\frac27T\) must be whole numbers, \(T\) must be a multiple of 28.

If \(T=28\), then the other players would have scored

\[
\frac{13}{28}(28)-15=13-15=-2,
\]

which is impossible.

Try the next multiple, \(T=56\):

\[
\frac{13}{28}(56)-15=26-15=11.
\]

This is possible, since 7 players can total 11 points while each scores at most 2 points.

The next multiple, \(T=84\), would give

\[
\frac{13}{28}(84)-15=39-15=24,
\]

but the 7 players can score at most

\[
7\cdot2=14.
\]

Therefore, their total was 11.

ANSWER 9: B

Problem 10:

The number \(QRS\) is divisible by 5. Since the digits are 1, 2, 3, 4, and 5, its last digit must be 5.

Therefore,

\[
S=5.
\]

Also, \(PQR\) is divisible by 4, so its last two digits \(QR\) must form a multiple of 4.

Using the remaining digits, the possible values of \(QR\) are

\[
12,\quad 24,\quad 32.
\]

Now use the condition that \(RST\) is divisible by 3.

If \(QR=12\), then \(R=2\). The remaining possibilities for \(T\) are 3 or 4:

\[
2+5+3=10,\qquad 2+5+4=11.