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 $5(2g-30)=g+210\implies 10g-150=g+210\implies 9g=360\implies g=\boxed{\textbf{(A)}\ 40}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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1950 AHSME Problems/Problem 30

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