1950 AHSME Problems/Problem 30

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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 $g = \boxed{\textbf{(A)}\ 40}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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