Difference between revisions of "2016 AMC 8 Problems/Problem 3"
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+ | == 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? | 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== |
+ | ===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 11:41, 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
:
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.