During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# 2020 AMC 8 Problems/Problem 13

Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?

$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

## Solution 1

After Jamal adds $x$ purple socks, he has $18+x$ purple socks and $6+18+12+x=x+36$ total socks, for a probability of drawing a purple sock of $$\dfrac{18+x}{36+x}=\dfrac{3}{5}.$$ Since $\dfrac{18+9}{36+9}=\dfrac{27}{45}=\dfrac35$, the answer is $\boxed{\textbf{(B) }9}$. ~icematrix

## Solution 2

The total number of socks that Jamal has is $\, 6+18+12=36$ socks. We are trying to determine how many purple socks he added to his drawer. Let's say he adds $x$ purple socks. This means that the total number of purple socks in his drawer will be $(18+x)$ and the new total number of socks in his drawer will be $(36+x)$. The ratio of purple socks to total socks in his drawer is now $\frac{60}{100}=\frac{3}{5}$. This leads to the equation $\frac{18+x}{36+x}=\frac{3}{5}$. Cross multiplying this equation gives us $90+5x=108+3x \implies 2x=18 \implies x=9$. Thus, Jamal added 9 purple socks $\implies\boxed{\textbf{(B) }9}$.
~junaidmansuri

## Solution 3

Let Jamal add $x$ more purple socks. Then we are told that $\frac{18+x}{6+18+12+x}=\frac35$. Cross multiplying and simplifying tells us that $x=\textbf{(B)}9$.

-franzliszt

~savannahsolver