2016 AMC 8 Problems/Problem 3: Difference between revisions

Revision as of 00:31, 24 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

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 7: / Line 7: @@
 ==Solutions==
 ===Solution 1===
-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>.
+Let <math>r</math> be the remaining student's score.  We 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===

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