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

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==Solutions==
 
==Solutions==
 
===Solution 1===
 
===Solution 1===
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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>.
+
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===
 
===Solution 2===

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

See Also

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

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