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

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==Video Solution==
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https://youtu.be/R2jD3a5SXAY?si=q-T8ZTrHYIb7j-xB
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A solution so simple a 12-year-old made it!
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~Elijahman~
  
 
==Video Solution (THINKING CREATIVELY!!!)==
 
==Video Solution (THINKING CREATIVELY!!!)==

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

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