1994 AHSME Problems/Problem 16


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?

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


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

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS