2008 AMC 8 Problems/Problem 20

Problem

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution

Let $b$ be the number of boys and $g$ be the number of girls.

$\frac23 b = \frac34 g \Rightarrow b = \frac98 g$

For $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\boxed{\textbf{(B)}\ 17}$ .

Solution 2

We know that $\frac23 B = \frac34 G$ or $\frac69 B = \frac68 G$ . So, the ratio of the number of boys to girls is 8:9 (Not 9:8 because the numbers 8 and 9 are in the denominators of the fractions). The smallest total number of students is $9 + 8 = \boxed{\textbf{(B)}\ 17}$ . ~DY, edited by SuperVince1

Video Solution by WhyMath

https://youtu.be/HseEFQDuh_c

~savannahsolver

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2008 AMC 8 Problems/Problem 20

Contents

Problem

Solution

Solution 2

Video Solution by WhyMath

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