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

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==Problem==
 
Jamal has a drawer containing <math>6</math> green socks, <math>18</math> purple socks, and <math>12</math> orange socks. After adding more purple socks, Jamal noticed that there is now a <math>60\%</math> chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
 
Jamal has a drawer containing <math>6</math> green socks, <math>18</math> purple socks, and <math>12</math> orange socks. After adding more purple socks, Jamal noticed that there is now a <math>60\%</math> chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
  
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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=x+36</math> total socks, for a probability of drawing a purple sock of <cmath>\dfrac{18+x}{36+x}=\dfrac{3}{5}.</cmath> Since <math>\dfrac{18+9}{36+9}=\dfrac{27}{45}=\dfrac35</math>, the answer is <math>\boxed{\textbf{(B) }9}</math>. ~icematrix
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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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==Solution 2==
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==Solution 2 (variant of Solution 1)==
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The total number of socks that Jamal has is <math>\, 6+18+12=36</math> socks. We are trying to determine how many purple socks he added to his drawer. Let's say he adds <math>x</math> purple socks. This means that the total number of purple socks in his drawer will be <math>(18+x)</math> and the new total number of socks in his drawer will be <math>(36+x)</math>. The ratio of purple socks to total socks in his drawer is now <math>\frac{60}{100}=\frac{3}{5}</math>. This leads to the equation <math>\frac{18+x}{36+x}=\frac{3}{5}</math>. Cross multiplying this equation gives us <math>90+5x=108+3x \implies 2x=18 \implies x=9</math>. Thus, Jamal added 9 purple socks <math>\implies\boxed{\textbf{(B) }9}</math>.<br>
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As in Solution 1, we have the equation <math>\frac{18+x}{36+x}=\frac{3}{5}</math>. Cross-multiplying yields <math>90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9</math>. Thus, Jamal added <math>\boxed{\textbf{(B) }9}</math> purple socks.
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~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
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==Solution 3==
 
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Let Jamal add <math>x</math> more purple socks. Then we are told that <math>\frac{18+x}{6+18+12+x}=\frac35</math>.  Cross multiplying and simplifying tells us that <math>x=\textbf{(B)}9</math>.
 
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-franzliszt
 
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/x9Di0yxUqeU
 
https://youtu.be/x9Di0yxUqeU
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~savannahsolver
 
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|num-b=12|num-a=14}}
 
{{AMC8 box|year=2020|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:57, 20 November 2020

Problem

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.

Video Solution

https://youtu.be/x9Di0yxUqeU

See also

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

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