Difference between revisions of "2020 AMC 10A Problems/Problem 2"

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==Video Solution==
 
==Video Solution==
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Education, The Study of Everything
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https://youtu.be/e8Qfe5GpEUg
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https://youtu.be/WUcbVNy2uv0
 
https://youtu.be/WUcbVNy2uv0
  

Revision as of 10:40, 7 November 2020

Problem

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$. What is the average of $a$ and $b$?

$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$

Solution

The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15$. Solving for $a+b$, we get $a+b=60$. Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{\textbf{(C) }30}$.

Video Solution

Education, The Study of Everything

https://youtu.be/e8Qfe5GpEUg

https://youtu.be/WUcbVNy2uv0

~IceMatrix

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

https://youtu.be/zVppmKOvx_w

~savannahsolver

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AMC 10 Problems and Solutions

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