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

(Solution 2)
Line 17: Line 17:
 
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
 
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
  
 +
==See Also==
 
{{AMC8 box|year=2016|num-b=2|num-a=4}}
 
{{AMC8 box|year=2016|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:30, 21 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)

See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png