Difference between revisions of "2016 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math> | <math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math> | ||
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| − | + | ==Solution 1== | |
Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | ||
| − | + | ==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== | ||
| + | https://youtu.be/ysYNhG_uG-I | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/TkZvMa30Juo?t=638 | ||
| + | |||
| + | ~pi_is_3.14 | ||
| + | |||
| + | ==See Also== | ||
{{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 17:04, 25 June 2025
Contents
Problem
Suppose that 
means 
What is the value of 
if
![\[2 * (5 * x)=1\]](http://latex.artofproblemsolving.com/c/2/f/c2f6515e388e34fda66b684dcf2dc3b67ef1bc83.png)
Solution 1
Let us plug in 
into 
. Thus it would be 
. Now we have 
. Plugging 
into , we have 
. Solving for we have ![\[-9+x=1\]](http://latex.artofproblemsolving.com/c/6/1/c61c1f668c35a5c27c69301b25294456176e3130.png)
![\[x=\boxed{\textbf{(D)} \, 10}\]](http://latex.artofproblemsolving.com/7/9/e/79ee27e8c64bd1f7b4906b292dcb3704d68e7c15.png)
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. 
