1994 AHSME Problems/Problem 16

Problem

Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?

$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71$

Solution

Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So $\frac{r-1}{r+b-1}=\frac{1}{7}.$ When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total marbles left in the bag. So $\frac{r}{r+b-2}=\frac{1}{5}.$ Cross multiplying for each yields $\begin{align*}7r-7=r+b-1&\implies 7r-6=r+b\\ 5r=r+b-2&\implies 5r+2=r+b.\end{align*}$ We can equate each of these expressions to yields $7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.$ Therefore, the total number of marbles is $r+b=4+18=\boxed{\textbf{(B) }22.}$

--Solution by TheMaskedMagician

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1994 AHSME Problems/Problem 16

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