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

Revision as of 23:22, 25 December 2022

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

Let $r$ be the remaining student's score. We 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)

https://youtu.be/EuAzkusSbpY

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=772

~ pi_is_3.14

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 20: / Line 20: @@
 ~savannahsolver
+== Video Solution by OmegaLearn ==
+https://youtu.be/51K3uCzntWs?t=772
+~ pi_is_3.14
 ==See Also==
 {{AMC8 box|year=2016|num-b=2|num-a=4}}
 {{MAA Notice}}

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