🔬 Model Lab

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B

Problem 1:
We need the sum of each triplet to check if it equals 1.

A. (1/2, 1/3, 1/6). Sum = 1/2+1/3+1/6 = 3/6+2/6+1/6 = 6/6 = 1.

B. (2, -2, 1). Sum = 2-2+1 = 1.

C. (0.1, 0.3, 0.6). Sum = 0.1+0.3+0.6 = 1.0.

D. (1.1, -2.1, 1.0). Sum = 1.1-2.1+1.0 = 2.1-2.1 = 0. This is not 1.

E. (-3/2, -5/2, 5). Sum = -3/2-5/2+5 = -8/2+5 = -4+5 = 1.

Only D does not sum to 1.

ANSWER 1: D

Problem 2:
We interpret the choices as fractions with x = 7:
A. 6/x, B. 6/(x+1), C. 6/(x-1), D. x/6, E. (x+1)/6.

Substitute x = 7:
A. 6/7 ≈ 0.86
B. 6/(7+1) = 6/8 = 3/4 = 0.75
C. 6/(7-1) = 6/6 = 1
D. 7/6 ≈ 1.17
E. (7+1)/6 = 8/6 ≈ 1.33

For positive numerators, the smaller the denominator the larger the fraction, and vice versa. Among A,B,C, 6/8 is smallest. D and E are both >1, so larger.

Hence B is the smallest.

ANSWER 2: B

Problem 3:
We need prime factorization of 250.

250 = 25 × 10 = 5² × 2 × 5 = 2 × 5³.

So the prime factors listed with multiplicity are 2,5,5,5.

The two smallest are 2 and 5. Their sum is 2+5 = 7.

Note choice A 2 would be just the smallest, D 10 would be 5+5, but we want the two smallest.

ANSWER 3: C

Problem 4:
We are asked for total students taking the AMC 8 from the three math classes.

Miss Germain's class: 11 students
Mr. Newton's class: 8 students
Mrs. Young's class: 9 students

Total = 11+8+9 = 19+9 = 28.

This is straightforward addition. The total 17 is not an option, so all three counts are included.

ANSWER 4: C

Problem 5:
Let total marbles = T, with r red, g green, b blue, so T = r+g+b.

"All but 6 are red" means all except 6 are red, so the non-red marbles number 6:
g + b = 6.

"All but 8 are green" means non-green number 8:
r + b = 8.

"All but 4 are blue" means non-blue number 4:
r + g = 4.

Add all three equations:
(g+b)+(r+b)+(r+g) = 6+8+4
2(r+g+b) = 18
2T = 18
T = 9.

Check: r = T-6 = 3, g = T-8 = 1, b = T-4 = 5, and 3+1+5=9.

ANSWER 5: C

Problem 6:
Wire length available is 24 cm. A shape can be made if its perimeter is ≤ 24 cm; since we want to bend it to form the shape, typically perimeter = 24 exactly if using all, but at most 24 is needed. Here check:

Hexagon: Regular with side 5 cm. Perimeter = 6×5 = 30 cm. 30 > 24, so cannot make.

Square: Area 36 cm². Side = √36 = 6 cm. Perimeter = 4×6 = 24 cm. Exactly uses wire, so can make.

Right triangle: Legs 6 and 8 cm. Hypotenuse by Pythagoras = √(6²+8²) = √(36+64) = √100 = 10 cm. Perimeter = 6+8+10 = 24 cm. Exactly uses wire, so can make.

Thus square and triangle only.

ANSWER 6: D

Problem 7:
Let original price = P.

Sale is half price: sale price = P/2 = 0.5P.

Coupon gives 20% discount on sale price, so you pay 100%-20% = 80% of sale price:
final price = 0.80 × 0.5P = 0.40P.

That is 40% of original, so the reduction is 100%-40% = 60% off original.

Tempting wrong: A 10 would add discounts incorrectly, C 40 is the percent paid, not off.

ANSWER 7: D

Problem 8:
Distance ≈ speed × time, ignoring light travel time.

Speed of sound = 1088 ft/s, time = 10 s:
distance in feet = 1088×10 = 10880 ft.

1 mile = 5280 ft, so distance in miles = 10880/5280 = 1088/528.

528×2 = 1056, remainder 32, so = 2 + 32/528 ≈ 2.06 miles.

To nearest half-mile among 1, 1½, 2, 2½, 3, 2.06 is closest to 2.

ANSWER 8: C

Problem 9:
Positions in row: _ _ _ _ _. Requirement: boy at each end, girls in middle, i.e., B G G G B pattern.

Number of ways to place the 2 distinct boys on the 2 end spots: 2! = 2.

Number of ways to place the 3 distinct girls in the 3 middle spots: 3! = 3×2×1 = 6.

By counting principle, total arrangements = 2×6 = 12.

Choices A-D are too small; they forget permuting within genders.

ANSWER 9: E

Problem 10:
Interpret the expression as (3/8 + 7/8) ÷ (4/5).

First add numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 means multiply by 5/4:
(5/4) ÷ (4/5) = (5/4)×(5/4) = 25/16.

This matches choice B. Choice A 1, C 2 would be from mis-adding, etc.

ANSWER 10: B

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

---

### Problem 1:
**Restatement:** We are asked to identify which of the given sets of three numbers does NOT add up to $1$.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.
- **B.** $2 + (-2) + 1 = 0 + 1 = 1$.
- **C.** $0.1 + 0.3 + 0.6 = 1.0 = 1$.
- **D.** $1.1 + (-2.1) + 1.0 = -1.0 + 1.0 = 0 \neq 1$.
- **E.** $-\frac{3}{2} + \left(-\frac{5}{2}\right) + 5 = -\frac{8}{2} + 5 = -4 + 5 = 1$.

The triplet in choice D has a sum of $0$, not $1$.

ANSWER 1: D

---

### Problem 2:
**Restatement:** Given $x = 7$, we want to find which of the expressions has the smallest value.

**Step-by-step evaluation of each choice:**
- **A.** $\frac{6}{x} = \frac{6}{7} \approx 0.857$
- **B.** $\frac{6}{x+1} = \frac{6}{7+1} = \frac{6}{8} = \frac{3}{4} = 0.75$
- **C.** $\frac{6}{x-1} = \frac{6}{7-1} = \frac{6}{6} = 1$
- **D.** $\frac{x}{6} = \frac{7}{6} \approx 1.167$
- **E.** $\frac{x+1}{6} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3} \approx 1.333$

Comparing the values, $0.75 < 0.857 < 1 < 1.167 < 1.333$. The smallest value is $\frac{6}{x+1} = \frac{3}{4}$.

ANSWER 2: B

---

### Problem 3:
**Restatement:** Find the sum of the two smallest prime factors of $250$.

**Step-by-step solution:**
1. Find the prime factorization of $250$:
   $$250 = 2 \times 125 = 2 \times 5^3$$
2. The prime factors of $250$ are $2$ and $5$.
3. The two smallest prime factors are therefore $2$ and $5$.
4. Their sum is:
   $$2 + 5 = 7$$

ANSWER 3: C

---

### Problem 4:
**Restatement:** We are given the number of AMC 8 participants in each of three teachers' classes:
- Miss Germain: $11$ students
- Mr. Newton: $8$ students
- Mrs. Young: $9$ students

**Step-by-step solution:**
To find the total number of students taking the contest, sum the students from all three classes:
$$\text{Total} = 11 + 8 + 9 = 28$$

ANSWER 4: C

---

### Problem 5:
**Restatement:** In a jar containing only red ($R$), green ($G$), and blue ($B$) marbles:
- All but $6$ are red $\implies G + B = 6$
- All but $8$ are green $\implies R + B = 8$
- All but $4$ are blue $\implies R + G = 4$

We need to find the total number of marbles, $T = R + G + B$.

**Step-by-step solution:**
Add the three equations together:
$$(G + B) + (R + B) + (R + G) = 6 + 8 + 4$$
$$2R + 2G + 2B = 18$$
$$2(R + G + B) = 18$$
Divide both sides by $2$:
$$R + G + B = 9$$

There are $9$ marbles in the jar.

ANSWER 5: C

---

### Problem 6:
**Restatement:** Haruki has a piece of wire $24\text{ cm}$ long. We must check which shapes can be formed with this wire (i.e., which shapes have a perimeter of exactly $24\text{ cm}$).

**Step-by-step check of each shape:**
1. **Regular hexagon with side length $5\text{ cm}$:**
   $$\text{Perimeter} = 6 \times 5 = 30\text{ cm} \neq 24\text{ cm}$$
   He cannot make the hexagon.

2. **Square of area $36\text{ cm}^2$:**
   $$\text{Side length} = \sqrt{36} = 6\text{ cm}$$
   $$\text{Perimeter} = 4 \times 6 = 24\text{ cm}$$
   He can make the square.

3. **Right triangle with legs $6\text{ cm}$ and $8\text{ cm}$:**
   $$\text{Hypotenuse} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$$
   $$\text{Perimeter} = 6 + 8 + 10 = 24\text{ cm}$$
   He can make the right triangle.

Thus, Haruki can make the square and triangle only.

ANSWER 6: D

---

### Problem 7:
**Restatement:** An item is on sale for half price ($50\%$ off), and an additional coupon gives a $20\%$ discount on the sale price. What percentage off the original price is the final price?

**Step-by-step solution:**
1. Let the original price be $P$.
2. The sale price is half of the original:
   $$\text{Sale Price} = 0.50P$$
3. The coupon reduces the sale price by $20\%$, meaning the customer pays $80\%$ of the sale price:
   $$\text{Final Price} = 0.80 \times (0.50P) = 0.40P$$
4. Since the customer pays $40\%$ of the original price, the total discount off the original price is:
   $$100\% - 40\% = 60\%$$

ANSWER 7: D

---

### Problem 8:
**Restatement:** Snoopy hears thunder $10$ seconds after seeing lightning. Sound travels at $1088\text{ ft/s}$, and $1\text{ mile} = 5280\text{ ft}$. Estimate the distance to the nearest half-mile.

**Step-by-step solution:**
1. Calculate the distance traveled by the sound in feet:
   $$\text{Distance} = 1088\text{ ft/s} \times 10\text{ s} = 10{,}880\text{ feet}$$
2. Convert this distance to miles:
   $$\frac{10{,}880}{5280} \approx 2.0606\text{ miles}$$
3. Round to the nearest half-mile (choices are $1$, $1\frac{1}{2}$, $2$, $2\frac{1}{2}$, $3$):
   - Distance to $2$: $|2.0606 - 2| = 0.0606$
   - Distance to $2.5$: $|2.0606 - 2.5| = 0.4394$

The closest value is $2$ miles.

ANSWER 8: C

---

### Problem 9:
**Restatement:** Five students (2 boys and 3 girls) are arranged in a row of 5 seats such that there is a boy at each end and the 3 girls are in the middle 3 seats. Find the number of possible arrangements.

**Step-by-step solution:**
The arrangement of the 5 positions must look like:
$$\text{Boy}_1, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Girl}, \quad \text{Boy}_2$$

1. **Placing the boys at the ends:**
   There are $2$ boys for the first seat and $1$ remaining boy for the last seat:
   $$2! = 2 \times 1 = 2\text{ ways}$$

2. **Placing the girls in the middle:**
   There are $3$ girls to arrange in the middle 3 seats:
   $$3! = 3 \times 2 \times 1 = 6\text{ ways}$$

3. **Total arrangements:**
   $$\text{Total} = 2 \times 6 = 12\text{ ways}$$

ANSWER 9: E

---

### Problem 10:
**Restatement:** Evaluate the complex fraction:
$$\frac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}$$

**Step-by-step solution:**
1. Simplify the numerator:
   $$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$$

2. Divide the numerator by the denominator:
   $$\frac{\frac{5}{4}}{\frac{4}{5}} = \frac{5}{4} \times \frac{5}{4} = \frac{25}{16}$$

ANSWER 10: B

# Problem 1:
We need the triplet whose sum is NOT 1. Check each:

- **A:** 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓
- **B:** 2 + (−2) + 1 = 1 ✓
- **C:** 0.1 + 0.3 + 0.6 = 1.0 ✓
- **D:** 1.1 + (−2.1) + 1.0 = 0 ✗ (This is NOT 1)
- **E:** −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✓

Only choice D fails to sum to 1.

ANSWER 1: D

# Problem 2:
Substitute x = 7 into each expression:

- **A:** 6/x = 6/7 ≈ 0.857
- **B:** 6/(x+1) = 6/8 = 0.75
- **C:** 6/(x−1) = 6/6 = 1
- **D:** x/6 = 7/6 ≈ 1.167
- **E:** (x+1)/6 = 8/6 ≈ 1.333

Comparing: 0.75 < 0.857 < 1 < 1.167 < 1.333. The smallest is 6/8 = 0.75, choice B.

ANSWER 2: B

# Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

The prime factors are 2 and 5. The two smallest prime factors are 2 and 5.

Their sum: 2 + 5 = 7.

(Trap: choice D is 10 = 2 × 5, the *product*, not the sum.)

ANSWER 3: C

# Problem 4:
The students taking the contest are in three classes:
11 (Germain) + 8 (Newton) + 9 (Young) = 28 students.

(Trap: choice E, 30, would come from miscounting, e.g., using 11 + 10 + 9.)

ANSWER 4: C

# Problem 5:
Let T = total marbles. "All but 6 are red" means red = T − 6. Similarly green = T − 8, blue = T − 4.

Since red + green + blue = T:
(T − 6) + (T − 8) + (T − 4) = T
3T − 18 = T
2T = 18
T = 9.

Check: red = 3, green = 1, blue = 5. Then "all but 6 are red" ✓ (1+5=6 non-red), "all but 8 are green" ✓ (3+5=8), "all but 4 are blue" ✓ (3+1=4). Total = 9. ✓

ANSWER 5: C

# Problem 6:
The wire is 24 cm, so each shape's perimeter must be ≤ 24 (exactly 24 to use the whole wire, or check against the wire length).

- **Hexagon:** 6 sides × 5 cm = 30 cm > 24. **Cannot** be made.
- **Square:** Area 36 cm² → side = √36 = 6 cm → perimeter = 4 × 6 = 24 cm ≤ 24. **Can** be made.
- **Right triangle:** Legs 6 and 8 → hypotenuse = √(36 + 64) = √100 = 10 cm → perimeter = 6 + 8 + 10 = 24 cm ≤ 24. **Can** be made.

Only the square and triangle work.

ANSWER 6: D

# Problem 7:
Let the original price be P.

Step 1: Half price → P/2 (this is 50% off).
Step 2: Coupon takes 20% off the *sale price*, so you pay 80% of P/2:
(0.8)(P/2) = 0.4P.

So the final price is 40% of the original, meaning the discount is 100% − 40% = **60% off**.

(Trap: choice C, 40, is the final price percentage; the question asks the discount. Trap: 50+20 = 70 is wrong because the 20% applies to the reduced price.)

ANSWER 7: D

# Problem 8:
Distance = speed × time = 1088 ft/sec × 10 sec = 10,880 feet.

Convert to miles: 10,880 ÷ 5280 ≈ 2.06 miles.

To the nearest half-mile: 2.06 is closer to 2 than to 2½ (since 2.25 is the midpoint, and 2.06 < 2.25).

ANSWER 8: C

# Problem 9:
The seating is: Boy _ _ Girl _ Girl _ Boy, with the boys at the two ends.

- The two end positions: choose which boy is on the left and which on the right: 2! = 2 ways.
- The three middle positions: arrange the three girls: 3! = 3 × 2 × 1 = 6 ways.

Total arrangements: 2 × 6 = 12.

(Trap: choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls can be permuted.)

ANSWER 9: E

# Problem 10:
Interpret the expression as the fraction with numerator 3/8 + 7/8 over denominator 4/5:

$$\frac{\frac{3}{8}+\frac{7}{8}}{\frac{4}{5}}$$

Numerator: 3/8 + 7/8 = 10/8 = 5/4.

Divide by 4/5 (multiply by 5/4):

(5/4) × (5/4) = 25/16.

Check: 25/16 = 1.5625, matching choice B.

(Trap: adding everything or multiplying incorrectly gives the other fractions listed.)

ANSWER 10: B

Problem 1:
We need the triplet whose sum is NOT 1.

- A: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✔
- B: 2 + (−2) + 1 = 1 ✔
- C: 0.1 + 0.3 + 0.6 = 1.0 ✔
- D: 1.1 + (−2.1) + 1.0 = (1.1 − 2.1) + 1.0 = −1.0 + 1.0 = 0 ✘
- E: −3/2 − 5/2 + 5 = −8/2 + 5 = −4 + 5 = 1 ✔

Only D fails.

ANSWER 1: D

Problem 2:
Substitute x = 7 into each expression:

- A: 6/x = 6/7 ≈ 0.857
- B: 6/(x+1) = 6/8 = 0.75
- C: 6/(x−1) = 6/6 = 1
- D: x/6 = 7/6 ≈ 1.167
- E: (x+1)/6 = 8/6 ≈ 1.333

Among fractions with numerator 6, the largest denominator gives the smallest value, so 6/8 = 0.75 beats 6/7 and 6/6. The last two are greater than 1, so they can't be smallest.

ANSWER 2: B

Problem 3:
Factor 250: 250 = 2 · 125 = 2 · 5³.

So the prime factors are 2 and 5 only. The two smallest (and only) prime factors are 2 and 5, and 2 + 5 = 7.

(Choice E, 12, would come from adding 2 + 10, but 10 isn't prime; choice D, 10, from 5 + 5, but we need two *different* prime factors.)

ANSWER 3: C

Problem 4:
Each teacher's class contributes its students; the classes are separate groups of students:

11 + 8 + 9 = 28.

ANSWER 4: C

Problem 5:
Let the total be T, with R red, G green, B blue, so R + G + B = T.

- "All but 6 are red" means the non-red marbles number 6: G + B = 6, so R = T − 6.
- "All but 8 are green": R + B = 8, so G = T − 8.
- "All but 4 are blue": R + G = 4, so B = T − 4.

Add: R + G + B = 3T − 18 = T, so 2T = 18 and T = 9.

Check: R = 3, G = 1, B = 5 → non-red = 6 ✔, non-green = 8 ✔, non-blue = 4 ✔.

ANSWER 5: C

Problem 6:
The wire has length 24 cm, so the shape's perimeter must be exactly 24 cm.

- Regular hexagon, side 5: perimeter = 6 · 5 = 30 cm. Too long — impossible.
- Square of area 36: side = √36 = 6, perimeter = 4 · 6 = 24 cm. ✔
- Right triangle with legs 6 and 8: hypotenuse = √(36 + 64) = 10, perimeter = 6 + 8 + 10 = 24 cm. ✔

So the square and triangle only. (Don't forget to include the hypotenuse in the triangle's perimeter!)

ANSWER 6: D

Problem 7:
Let the original price be 100.

Half price: 100 × 0.5 = 50.
Coupon takes 20% off the sale price: 50 × 0.8 = 40.

Final price is 40, which is 60 less than 100, i.e., 60% off.

(Choice E, 70, comes from wrongly adding 50% + 20%; choice C, 40, is the percentage *paid*, not the discount.)

ANSWER 7: D

Problem 8:
Distance sound traveled in 10 seconds: 1088 × 10 = 10,880 feet.

Convert to miles: 10,880 / 5280 ≈ 2.06 miles.

To the nearest half mile, 2.06 rounds to 2 miles.

ANSWER 8: C

Problem 9:
Seats: B _ _ _ B, with boys at the two ends and girls in the three middle seats.

- Boys in the two end seats: 2! = 2 ways.
- Girls in the three middle seats: 3! = 6 ways.

Total: 2 × 6 = 12.

(Choice D, 6, forgets the boys can swap; choice A, 2, forgets the girls' orderings.)

ANSWER 9: E

Problem 10:
Compute the numerator first: 3/8 + 7/8 = 10/8 = 5/4.

Then divide by 4/5:
(5/4) ÷ (4/5) = (5/4) · (5/4) = 25/16.

ANSWER 10: B

Problem 1:

We need find which triplet does **not** have sum \(1\).

- A: \(\frac12+\frac13+\frac16=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=1\)
- B: \(2+(-2)+1=1\)
- C: \(0.1+0.3+0.6=1.0\)
- D: \(1.1+(-2.1)+1.0=1.1-2.1+1.0=0\), not \(1\)
- E: \(-\frac32-\frac52+5=-\frac82+5=-4+5=1\)

Only choice D does not add to \(1\).

ANSWER 1: D

Problem 2:

We substitute \(x=7\) into each expression:

- A: \(6^x=6^7\), a very large number.
- B: \(6^{x+1}=6^8\), even larger.
- C: \(6^{x-1}=6^6=46656\).
- D: \(x^6=7^6=117649\).
- E: \(x^{1/6}=7^{1/6}\), the sixth root of \(7\).

Since \(7^{1/6}\) is a little more than \(1\), it is much smaller than the large powers in A–D.

ANSWER 2: E

Problem 3:

Factor \(250\):

\[
250=25\cdot 10=5^2\cdot 2\cdot 5=2\cdot 5^3.
\]

The prime factors are \(2\) and \(5\). The two smallest prime factors are therefore \(2\) and \(5\).

\[
2+5=7.
\]

Choices A and B are individual prime factors, while D and E do not equal the required sum.

ANSWER 3: C

Problem 4:

Add the numbers of students from the three classes:

\[
11+8+9=19+9=28.
\]

Thus, \(28\) mathematics students are taking the AMC 8.

The nearby choices \(27\) and \(29\) would result from an addition error.

ANSWER 4: C

Problem 5:

Let the total number of marbles be \(N\).

- “All but 6 are red” means there are \(N-6\) red marbles.
- “All but 8 are green” means there are \(N-8\) green marbles.
- “All but 4 are blue” means there are \(N-4\) blue marbles.

Since every marble is exactly one color,

\[
(N-6)+(N-8)+(N-4)=N.
\]

So,

\[
3N-18=N
\]

\[
2N=18
\]

\[
N=9.
\]

Check: there are \(3\) red, \(1\) green, and \(5\) blue marbles, for a total of \(9\). Indeed, all but \(6\) are red, all but \(8\) are green, and all but \(4\) are blue.

ANSWER 5: C

Problem 6:

Haruki has \(24\) cm of wire, so each shape must have perimeter \(24\) cm.

- Regular hexagon with side length \(5\):

\[
6\cdot 5=30\text{ cm}.
\]

He cannot make the hexagon because \(30>24\).

- Square with area \(36\text{ cm}^2\):

\[
\text{side length}=\sqrt{36}=6\text{ cm}.
\]

Its perimeter is

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

He can make the square.

- Right triangle with legs \(6\) and \(8\):

Its hypotenuse is

\[
\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.
\]

Its perimeter is

\[
6+8+10=24\text{ cm}.
\]

He can make the triangle.

Thus, he can make the square and triangle only.

ANSWER 6: D

Problem 7:

“Half price” means the sale price is \(50\%\) of the original price.

Then the coupon takes another \(20\%\) off the sale price. Thus the customer pays

\[
80\%\text{ of }50\%=0.80\cdot 0.50=0.40=40\%
\]

of the original price.

So the discount is

\[
100\%-40\%=60\%.
\]

A tempting error is to add \(50\%+20\%=70\%\), but the \(20\%\) discount is applied to the already reduced sale price, not the original price.

ANSWER 7: D

Problem 8:

The thunder takes \(10\) seconds to travel from the lightning to Snoopy.

Distance in feet:

\[
1088\cdot 10=10880\text{ feet}.
\]

Convert to miles:

\[
\frac{10880}{5280}\approx 2.06\text{ miles}.
\]

The nearest half-mile to \(2.06\) is \(2\) miles.

It is much closer to \(2\) than to \(2\frac12\).

ANSWER 8: C

Problem 9:

There are two boys, and they must occupy the two end seats.

The boys can be arranged in the ends in

\[
2!=2
\]

ways.

The three girls fill the three middle seats in

\[
3!=6
\]

ways.

Therefore, the total number of arrangements is

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

Choices such as \(2\) or \(6\) count only the boys’ or girls’ arrangements, not both.

ANSWER 9: E

Problem 10:

The expression is

\[
\frac{\frac38+\frac78}{\frac45}.
\]

First add the fractions in the numerator:

\[
\frac38+\frac78=\frac{10}{8}=\frac54.
\]

Now divide by \(\frac45\):

\[
\frac54\div\frac45=\frac54\cdot\frac54=\frac{25}{16}.
\]

Therefore the value is \(\frac{25}{16}\).

ANSWER 10: B