2016 AMC 8 Problems/Problem 3: Difference between revisions
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==Solutions== | ==Solutions== | ||
===Solution 1=== | ===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=== | ===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>. | 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== | ==Video Solution== | ||
https:// | 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}} | {{AMC8 box|year=2016|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 14:57, 7 September 2025
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
