2016 AMC 8 Problems/Problem 3: Difference between revisions
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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>. | ||
{{AMC8 box|year=2016|num-b=2|num-a=4}} | |||
{{MAA Notice}} | |||
Revision as of 08:44, 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 | |
| 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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