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

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==Solution 1==
 
==Solution 1==
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After Jamal adds <math>x</math> purple socks, he has <math>(18+x)</math> purple socks and <math>6+18+12+x=(36+x)</math> total socks. This means the probability of drawing a purple sock is <math>\frac{18+x}{36+x}</math>, so we obtain <cmath>\frac{18+x}{36+x}=\frac{3}{5}</cmath> Since <math>\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}</math>, the answer is <math>\boxed{\textbf{(B) }9}</math>.
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After Jamal adds <math>x</math> purple socks, he has <math>(18+x)</math> purple socks and <math>6+18+12+x=(36+x)</math> total socks. This means the probability of drawing a purple sock is <math>\frac{18+x}{36+x}</math>, so we obtain <cmath>\frac{18+x}{36+x}=\frac{3}{5}</cmath> Since <math>\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{6}</math>, the answer is <math>\boxed{\textbf{(B) }9}</math>.
  
 
==Solution 2 (variant of Solution 1)==
 
==Solution 2 (variant of Solution 1)==

Revision as of 12:15, 17 January 2022

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 $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{6}$, 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 by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=u81EWYcC0Wg

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/x9Di0yxUqeU

~savannahsolver

Video Solution

https://youtu.be/xjwDsaRE_Wo

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=546

~Interstigation

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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