Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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− | + | Four students take an exam. Three of their scores are <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score? | |
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math> | <math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math> | ||
==Solution== | ==Solution== | ||
− | We | + | |
+ | We can call the remaining score <math>r</math>. We also know that the average, 70, is equal to <math>\frac{70 + 80 + 90 + r}{4}</math>. We can use basic algebra to solve for <math>r</math>: <cmath>\frac{70 + 80 + 90 + r}{4} = 70</cmath> <cmath>\frac{240 + r}{4} = 70</cmath> <cmath>240 + r = 280</cmath> <cmath>r = 40</cmath> giving us the answer of <math>\boxed{\textbf{(A)}\ 40}</math>. | ||
+ | |||
+ | ==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 <math>\boxed{\textbf{(A)}\ 40}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
+ | |||
+ | {{AMC8 box|year=2016|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 12:05, 5 July 2020
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 : giving us the answer of .
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 .
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.