# 2004 AMC 10B Problems/Problem 14

## Problem

A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only $\frac{1}{3}$ of the marbles in the bag are blue. Then yellow marbles are added to the bag until only $\frac{1}{5}$ of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

$\mathrm{(A) \ } \frac{1}{5} \qquad \mathrm{(B) \ } \frac{1}{4} \qquad \mathrm{(C) \ } \frac{1}{3} \qquad \mathrm{(D) \ } \frac{2}{5} \qquad \mathrm{(E) \ } \frac{1}{2}$

## Solution

We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed $\frac{1}{5}$ of the marbles in the bag. This means that there were $x$ blue and $4x$ other marbles, for some $x$. When we double the number of blue marbles, there will be $2x$ blue and $4x$ other marbles, hence blue marbles now form $\boxed{\mathrm{(C)\ }\frac{1}{3}}$ of all marbles in the bag.

## Solution 2

Assume that there are $2$ blue marbles and $1$ red marble. You don't have to add any red marbles to make the quantity $\frac{1}{3}$. Then, to make the blue marbles $\frac{1}{5}$, add $7$ yellow marbles. Doubling the blue marbles makes $4$ blue marbles out of $12$ total marbles, or $\boxed{\mathrm{(C)\ }\frac{1}{3}}$ of all marbles in the bag.

## Solution 3

If there are initially $B$ blue marbles in the bag, after red marbles are added, then the total number of marbles in the bag must be $3B$. Then after the yellow marbles are added, the number of marbles in the bag must be $5B$. Finally, adding $B$ blue marbles to the bag gives $2B$ blue marbles out of $6B$ total marbles. Thus $\boxed{1/3}$ of the marbles are blue.

## Solution 4

Just before the number of blue marbles is doubled, the ratio of blue marbles to non-blue marbles is 1 to 4. Doubling the number of blue marbles makes the ratio 2 to 4, so $\dfrac{2}{2+4} = \boxed{1/3}$ of the marbles are blue.