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Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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==Solution==
 
==Solution==
  
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We can call the remaining score <math>r</math>.  We also know that the average, 70, is equal to <math>\frac{70 + 80 + 90 + r}{4}</math>.  We can use basic algebra to solve for <math>r</math>: <cmath>\frac{70 + 80 + 90 + r}{4} = 70</cmath> <cmath>\frac{240 + r}{4} = 70</cmath> <cmath>240 + r = 280</cmath> <cmath>r = 40</cmath> giving us the answer of <math>\boxed{\text{A} (40)}</math>.
+
We can call the remaining score <math>r</math>.  We also know that the average, 70, is equal to <math>\frac{70 + 80 + 90 + r}{4}</math>.  We can use basic algebra to solve for <math>r</math>: <cmath>\frac{70 + 80 + 90 + r}{4} = 70</cmath> <cmath>\frac{240 + r}{4} = 70</cmath> <cmath>240 + r = 280</cmath> <cmath>r = 40</cmath> giving us the answer of <math>\boxed{\textbf{(A)}\ 40}</math>.
  
 
{{AMC8 box|year=2016|num-b=2|num-a=4}}
 
{{AMC8 box|year=2016|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:04, 28 November 2016

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 $\boxed{\textbf{(A)}\ 40}$.

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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