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2020 AMC 8 Problems/Problem 13

Revision as of 17:17, 19 November 2020 by Savannahsolver (talk | contribs)

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 he adds $x$ purple socks, the probability becomes \[\frac{18+x}{6+18+12+x}\implies\frac{18+x}{36+x}=\frac{3}{5}.\] Then, the answer is $\textbf{(B) }9$ because $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$. ~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

Video Solution

https://youtu.be/x9Di0yxUqeU

~savannahsolver

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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
All AJHSME/AMC 8 Problems and Solutions

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