Difference between revisions of "2016 AMC 8 Problems/Problem 10"
m (→Solution) |
|||
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| − | 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 < | + | 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> |
{{AMC8 box|year=2016|num-b=9|num-a=11}} | {{AMC8 box|year=2016|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:23, 27 November 2016
Suppose that 
means 
What is the value of 
if
![\[2 * (5 * x)=1\]](http://latex.artofproblemsolving.com/c/2/f/c2f6515e388e34fda66b684dcf2dc3b67ef1bc83.png)
Solution
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)
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
