Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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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>. | 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>. | ||
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==Solution 2== | ==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>. | 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 11:04, 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 .