# 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 $$r+b=4+18=\boxed{\textbf{(B) }22.}$$