Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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| + | ==Solution 2== | ||
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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>. | ||
Revision as of 09:57, 24 October 2018
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 can call the remaining score 
. We also know that the average, 70, is equal to 
. We can use basic algebra to solve for : ![\[\frac{70 + 80 + 90 + r}{4} = 70\]](http://latex.artofproblemsolving.com/f/6/4/f6415bdf4f17abab730096d189e087248be16df0.png)
![\[\frac{240 + r}{4} = 70\]](http://latex.artofproblemsolving.com/8/7/2/872a61211590971bc10753d0b06a09429d6b659f.png)
![\[240 + r = 280\]](http://latex.artofproblemsolving.com/d/9/c/d9c40d53447be6e76479604c4397300d6f031ba6.png)
![\[r = 40\]](http://latex.artofproblemsolving.com/c/a/a/caa81b049ae1aead94bf9e9360c38d1c9da0f084.png)
giving us the answer of 
.
| 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. 
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 .
