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 \[2g-30=b\] \[5b=g+210\]

Substituting $b$ for $2g-30$, we get $5(2g-30)=g+210\implies 10g-150=g+210\implies 9g=360\implies g=\boxed{\textbf{(A)}\ 40}$

See Also

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