Difference between revisions of "2016 AMC 8 Problems/Problem 10"
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==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 <cmath>2*(15-x)</cmath>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 <cmath>2*(15-x)</cmath>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> | ||
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+ | {{AMC8 box|year=2016|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Revision as of 09:46, 23 November 2016
Suppose that means What is the value of if
Solution
Let us plug in into . Thus it would be . Now we have . Plugging into . We have . Solving for we have
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AJHSME/AMC 8 Problems and Solutions |
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