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

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==Solution 3 (non-algebraic)==
 
==Solution 3 (non-algebraic)==
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<math>6</math> green socks and <math>12</math> orange socks together should be <math>100\%-60\% = 40\%</math> of the new total number of socks, so that new total must be <math>\frac{6+12}{0.4}= 45</math>. It follows that <math>45-6-18-12=\boxed{\textbf{(B) }9}</math> purple socks were added.
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<math>6</math> green socks and <math>12</math> orange socks together should be <math>100\%-60\% = 40\%</math> of the new total number of socks, so that new total must be <math>\frac{6+12}{0.4}= 45</math>. Therefore, <math>45-6-18-12=\boxed{\textbf{(B) }9}</math> purple socks were added.
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=bHNrBwwUCMI
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 +
~NiuniuMaths
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/UnVo6jZ3Wnk?si=6Ux3fNTMn5tSUnP3&t=1833
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 +
~Math-X
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==Video Solution (🚀Very Quick🚀)==
 +
https://youtu.be/0nnyYRYi-40
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by North America Math Contest Go Go Go==
 +
 
 +
https://www.youtube.com/watch?v=u81EWYcC0Wg
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 +
~North America Math Contest Go Go Go
 +
 
 +
==Video Solution by WhyMath==
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https://youtu.be/x9Di0yxUqeU
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 +
~savannahsolver
  
 
==Video Solution==
 
==Video Solution==
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https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)
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https://youtu.be/xjwDsaRE_Wo
  
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==Video Solution by Interstigation==
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https://youtu.be/YnwkBZTv5Fw?t=546
  
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https://youtu.be/xjwDsaRE_Wo
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~Interstigation
  
 
==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}}

Latest revision as of 15:34, 26 January 2024

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.

Video Solution by NiuniuMaths (Easy to understand!)

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

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=6Ux3fNTMn5tSUnP3&t=1833

~Math-X

Video Solution (🚀Very Quick🚀)

https://youtu.be/0nnyYRYi-40

~Education, the Study of Everything

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
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All AJHSME/AMC 8 Problems and Solutions

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