Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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==Solution== | ==Solution== | ||
We see that <math>80-70=10</math> and <math>90-70=20</math>. We then find that <math>10+20=30</math>. We want our average to be <math>70</math>, so we find <math>70-30=40</math>. So our final answer is <math>\boxed{\textbf{(A) }40}</math>. | We see that <math>80-70=10</math> and <math>90-70=20</math>. We then find that <math>10+20=30</math>. We want our average to be <math>70</math>, so we find <math>70-30=40</math>. So our final answer is <math>\boxed{\textbf{(A) }40}</math>. | ||
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+ | {{AMC8 box|year=2016|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 09:44, 23 November 2016
3. Four students take an exam. Three of their scores are and . If the average of their four scores is , then what is the remaining score?
Solution
We see that and . We then find that . We want our average to be , so we find . So our final answer is .
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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