Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (added .)
(→‎Video Solution)
 
(21 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
3. Four students take an exam. Three of their scores are <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score?
+
== Problem ==
 +
 
 +
Four students take an exam. Three of their scores are <math>70, 80,</math> and <math>90</math>. If the average of their four scores is <math>70</math>, then what is the remaining score?
  
 
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math>
 
<math>\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70</math>
  
−
==Solution==
+
==Solutions==
−
We see that <math>80-70=10</math> and <math>90-70=20</math>.  We then find that <math>10+20=30</math>.  We want our average to be <math>70</math>, so we find <math>70-30=40</math>. So our final answer is <math>\boxed{\textbf{(A) }40}</math>.
+
===Solution 1===
 +
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===
 +
 
 +
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
 +
https://youtu.be/R2jD3a5SXAY?si=brG-V2T2JYRkh_qC
 +
A solution so simple a 12-year-old made it!
 +
~Elijahman~
 +
 
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/R2jD3a5SXAY?si=q-T8ZTrHYIb7j-xB
 +
 
 +
A solution so simple a 12-year-old made it!
 +
 
 +
~Elijahman~
 +
 
 +
==Video Solution (THINKING CREATIVELY!!!)==
 +
https://youtu.be/jRPgMzBXYLc
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/EuAzkusSbpY
 +
 
 +
~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}}

Latest revision as of 10:41, 20 June 2024

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://youtu.be/R2jD3a5SXAY?si=brG-V2T2JYRkh_qC A solution so simple a 12-year-old made it! ~Elijahman~


Video Solution

https://youtu.be/R2jD3a5SXAY?si=q-T8ZTrHYIb7j-xB

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/jRPgMzBXYLc

~Education, the Study of Everything

Video Solution

https://youtu.be/EuAzkusSbpY

~savannahsolver

Video Solution by OmegaLearn

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

~ pi_is_3.14


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_3&oldid=220827"