Difference between revisions of "2016 AMC 8 Problems/Problem 3"
(→Solution 2) |
(→Solution 2) |
||
Line 11: | Line 11: | ||
===Solution 2=== | ===Solution 2=== | ||
− | Since <math>90</math> is <math>20</math> more than <math>70</math> and <math>80</math> is <math>10</math> more than <math>70</math>, for <math>70</math> to be the average, the other number must be <math>30</math> less than <math>70</math>, or <math>\boxed{\textbf{(A)}\ 40}</math>. | + | Since <math>90</math> is <math>20</math> more than <math>70</math>, and <math>80</math> is <math>10</math> more than <math>70</math>, for <math>70</math> to be the average, the other number must be <math>30</math> less than <math>70</math>, or <math>\boxed{\textbf{(A)}\ 40}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 17:06, 1 April 2021
Problem
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?
Solutions
Solution 1
We can call the remaining score . We also know that the average, 70, is equal to . We can use basic algebra to solve for : giving us the answer of .
Solution 2
Since is more than , and is more than , for to be the average, the other number must be less than , or .
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
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.