# Difference between revisions of "1994 AHSME Problems/Problem 16"

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<math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71 </math> | <math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71 </math> | ||

==Solution== | ==Solution== | ||

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

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+ | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] |

## Latest revision as of 20:56, 20 July 2014

## 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?

## Solution

Let and be the number of red and blue marbles originally in the bag respectively. After red marble is removed, there are marbles left in the bag and red marbles left. So When blue marbles are removed, there are red marbles and total marbles left in the bag. So Cross multiplying for each yields We can equate each of these expressions to yields Therefore, the total number of marbles is

--Solution by TheMaskedMagician