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Difference between revisions of "2019 AMC 8 Problems/Problem 15"

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==Solution 1==
 
==Solution 1==
 
The number of people wearing caps and sunglasses is  
 
The number of people wearing caps and sunglasses is  
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<math>\frac{2}{5}\cdot35=14</math>. So then 14 people out of the 50 people wearing sunglasses also have caps.  
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<math>\frac{2}{5}\cdot35=14</math>. So then, 14 people out of the 50 people wearing sunglasses also have caps.
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<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math>
 
<math>\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}</math>
  

Revision as of 23:40, 19 December 2022

Problem 15

On a beach, $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is $\frac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap? $\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}$

Video Solution by OmegaLearn

https://youtu.be/6xNkyDgIhEE?t=252

~ pi_is_3.14

Solution 1

The number of people wearing caps and sunglasses is $\frac{2}{5}\cdot35=14$. So then, 14 people out of the 50 people wearing sunglasses also have caps.

$\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$

Solution Explained

https://youtu.be/gOZOCFNXMhE ~ The Learning Royal

Video Solution

https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx

Another video - https://www.youtube.com/watch?v=afMsUqER13c

https://youtu.be/37UWNaltvQo -Happytwin

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=omRgmX7KXOg&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=16

Video Solution

https://youtu.be/9nbaMSAQCNU

~savannahsolver

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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