Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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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>. | 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== | ==Video Solution== | ||
| − | https:// | + | 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 | https://youtu.be/EuAzkusSbpY | ||
~savannahsolver | ~savannahsolver | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/51K3uCzntWs?t=772 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==See Also== | ==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}} | ||
Latest revision as of 10:41, 20 June 2024
Contents
Problem
Four students take an exam. Three of their scores are 
and 
. If the average of their four scores is 
, then what is the remaining score?
Solutions
Solution 1
Let 
be the remaining student's score. We know that the average, 70, is equal to 
. We can use basic algebra to solve for : ![\[\frac{70 + 80 + 90 + r}{4} = 70\]](http://latex.artofproblemsolving.com/f/6/4/f6415bdf4f17abab730096d189e087248be16df0.png)
![\[\frac{240 + r}{4} = 70\]](http://latex.artofproblemsolving.com/8/7/2/872a61211590971bc10753d0b06a09429d6b659f.png)
![\[240 + r = 280\]](http://latex.artofproblemsolving.com/d/9/c/d9c40d53447be6e76479604c4397300d6f031ba6.png)
![\[r = 40\]](http://latex.artofproblemsolving.com/c/a/a/caa81b049ae1aead94bf9e9360c38d1c9da0f084.png)
giving us the answer of 
.
Solution 2
Since is 
more than , and 
is 
more than , for to be the average, the other number must be 
less than , or .
== 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!!!)
~Education, the Study of Everything
Video Solution
~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. 
