Difference between revisions of "2020 AMC 8 Problems/Problem 13"

Revision as of 15:45, 14 December 2020

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 $\frac{6+12}{0.4}= 45$ . It follows that $45-6-18-12=\boxed{\textbf{(B) }9}$ purple socks were added.

Video Solution

https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)

https://youtu.be/xjwDsaRE_Wo

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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