Difference between revisions of "2020 AMC 8 Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
− | 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}{ | + | 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)== | ||
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==Solution 3 (non-algebraic)== | ==Solution 3 (non-algebraic)== | ||
<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. | <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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+ | ==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== | ==Video Solution== | ||
https://youtu.be/xjwDsaRE_Wo | https://youtu.be/xjwDsaRE_Wo | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=546 | ||
+ | |||
+ | ~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}} |
Revision as of 13:15, 17 January 2022
Contents
Problem 13
Jamal has a drawer containing green socks, purple socks, and orange socks. After adding more purple socks, Jamal noticed that there is now a chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
Solution 1
After Jamal adds purple socks, he has purple socks and total socks. This means the probability of drawing a purple sock is , so we obtain Since , the answer is .
Solution 2 (variant of Solution 1)
As in Solution 1, we have the equation . Cross-multiplying yields . Thus, Jamal added purple socks.
Solution 3 (non-algebraic)
green socks and orange socks together should be of the new total number of socks, so that new total must be . It follows that 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
~savannahsolver
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.