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>. | ||
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+ | == Video Solution | ||
+ | https://youtu.be/R2jD3a5SXAY?si=brG-V2T2JYRkh_qC | ||
+ | A solution so simple a 12-year-old made it! | ||
+ | ~Elijahman~ | ||
+ | |||
==Video Solution (THINKING CREATIVELY!!!)== | ==Video Solution (THINKING CREATIVELY!!!)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== |
Revision as of 10:40, 20 June 2024
Contents
[hide]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 : 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 (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.