Difference between revisions of "2016 AMC 8 Problems/Problem 3"

Revision as of 18:06, 1 April 2021

Problem

Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?

$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$

Solutions

Solution 1

We can call the remaining score $r$ . We also know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ . We can use basic algebra to solve for $r$ : $\frac{70 + 80 + 90 + r}{4} = 70$ $\frac{240 + r}{4} = 70$ $240 + r = 280$ $r = 40$ giving us the answer of $\boxed{\textbf{(A)}\ 40}$ .

Solution 2

Since $90$ is $20$ more than $70$ , and $80$ is $10$ more than $70$ , for $70$ to be the average, the other number must be $30$ less than $70$ , or $\boxed{\textbf{(A)}\ 40}$ .

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.

@@ Line 11: / Line 11: @@
 ===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 <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==

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