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Difference between revisions of "2016 AMC 8 Problems/Problem 3"

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Since 90 is 20 more than 70 and 80 is ten more than 70, for 70 to be the average, the other number must be thirty less than 70, or  <math>\boxed{\textbf{(A)}\ 40}</math>.
 
Since 90 is 20 more than 70 and 80 is ten more than 70, for 70 to be the average, the other number must be thirty less than 70, or  <math>\boxed{\textbf{(A)}\ 40}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
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{{AMC8 box|year=2016|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 11:05, 5 July 2020

Four students take an exam. Three of their scores are $70, 80,$ and $90$. If the average of their four scores is $70$, then what is the remaining score?

$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$

Solution

We can call the remaining score $r$. We also know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$. We can use basic algebra to solve for $r$: \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{\textbf{(A)}\ 40}$.

Solution 2

Since 90 is 20 more than 70 and 80 is ten more than 70, for 70 to be the average, the other number must be thirty less than 70, or $\boxed{\textbf{(A)}\ 40}$.

Video Solution

https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AJHSME/AMC 8 Problems and Solutions

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