AMC 8 · 2008 · #20

Grade 6 rate-ratio

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Chained Ratio Linking

ratio-proportionfraction-arithmeticlcmmultiples identify-subproblems ↑ Prerequisites: fraction-arithmeticlcm

📏 Medium solution 💡 3 insights

Problem

In Mr. Neatkin's class,

\frac{2}{3}

of the boys passed a penmanship test and

\frac{3}{4}

of the girls passed. The number of boys who passed equals the number of girls who passed. Find the smallest possible total number of students in the class.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Set Up an Equation

The two fractions describe two counts that the problem says are equal — that is exactly an equation waiting to be written, so Tool #3 (Set Up an Equation) drives the work. Tool #4 (Introduce a Variable) gives us letters for the number of boys and the number of girls so we can write the equation. The word "minimum" is the cue for Tool #14 (Consider Extreme Cases): once the equation forces a ratio between boys and girls, the smallest whole-number pair fitting that ratio gives the answer.

1STEP 1

Let b be the number of boys and g the number of girls; then $\frac{2}{3}b$ boys and $\frac{3}{4}g$ girls passed.

b = number of boys, g = number of girls

2STEP 2

"Equal numbers passed" means the two passing counts are equal: $\frac{2}{3}b = \frac{3}{4}g$ .

\frac{2}{3}b = \frac{3}{4}g

3STEP 3

Multiply both sides by the LCM 12 to clear denominators, giving 8b = 9g.

12 ·

\frac{2}{3}b

= 12 ·

\frac{3}{4}g

→ 8b = 9g

4STEP 4

Since 8 and 9 are coprime, whole-number solutions force the ratio b : g = 9 : 8.

8b = 9g →

\frac{b}{g} = \frac{9}{8}

5STEP 5

The smallest pair is b = 9, g = 8; check: $\frac{2}{3}$ (9) = 6 boys and $\frac{3}{4}$ (8) = 6 girls passed — equal.

b = 9, g = 8 (passers: 6 = 6 ✓)

6STEP 6

Add: 9 boys + 8 girls = 17 students, the smallest possible class.

b + g = 9 + 8 = 17 → (B)

Answer

Check the next-smallest pair: b = 18, g = 16 also satisfies 8b = 9g, and gives

\frac{2}{3}

(18) = 12 boys and

\frac{3}{4}

(16) = 12 girls passing — equal, but the total is 34, larger than 17. Every valid pair is a whole-number multiple of (9, 8), so 17 is truly the minimum. The wrong choices line up with traps: (A) 12 ignores that b and g must be different sizes; (C) 24, (D) 27, (E) 36 are larger multiples of the ratio.

💡Key takeaway

When two fractions of two groups give the same count, write one equation, clear the fractions, and the smallest whole numbers in the resulting ratio give the answer — here 9 boys and 8 girls, total 17.

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