Difference between revisions of "2016 AMC 8 Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/cR1GDMq1Cv4?si=KktuZcPdJxqC83rg | ||
+ | |||
+ | A solution so simple that a 12-year-old made it! | ||
+ | |||
+ | ~Elijahman~ | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/0FhGMy0mCVU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
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{{AMC8 box|year=2016|num-b=9|num-a=11}} | {{AMC8 box|year=2016|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 18:04, 25 June 2025
Contents
Problem
Suppose that means
What is the value of
if
Solution 1
Let us plug in into
. Thus it would be
. Now we have
. Plugging
into
, we have
. Solving for
we have
Solution 2
Let us set a variable equal to
. Solving for y in the equation
, we see that y is equal to five. By substitution, we see that
= 5. Solving for x in the equation
we get
Video Solution
https://youtu.be/cR1GDMq1Cv4?si=KktuZcPdJxqC83rg
A solution so simple that a 12-year-old made it!
~Elijahman~
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=638
~pi_is_3.14
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.