2020 AMC 8 Problems/Problem 13

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 $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$, so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$, the answer is $\boxed{\textbf{(B) }9}$.

Solution 2 (variant of Solution 1)

As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$. Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$. Thus, Jamal added $\boxed{\textbf{(B) }9}$ purple socks.

Solution 3 (non-algebraic)

$6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$. Therefore, $45-6-18-12=\boxed{\textbf{(B) }9}$ purple socks were added.

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See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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