2008 AMC 8 Problems/Problem 20

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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 numerator, and the smallest possible value is $8$. This yields $9$ boys. The minimum number of students is $8+9=\boxed{\textbf{(B)}\ 17}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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