1950 AHSME Problems/Problem 30

Problem

From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:

$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$

Solution

Let us represent the number of boys $b$ , and the number of girls $g$ .

From the first sentence, we get that $2(g-15)=b$ .

From the second sentence, we get $5(b-45)=g-15$ .

Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$ .

Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , we get $g = \boxed{\textbf{(A)}\ 40}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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All AHSME Problems and Solutions

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1950 AHSME Problems/Problem 30

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