Difference between revisions of "2016 AMC 8 Problems/Problem 3"
Reaganchoi (talk | contribs) m (added .) |
|||
Line 5: | Line 5: | ||
==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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.