Difference between revisions of "2016 AMC 10B Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \ | + | What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \tfrac{1}{2}</math>? |
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math> | ||
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==Solution== | ==Solution== | ||
− | <math>\frac{\frac{1}{a} | + | Factorizing the numerator, <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> then becomes <math>\frac{\frac{5}{2}}{a^{2}}</math> which is equal to <math>\frac{5}{2}\cdot 2^2</math> which is <math>\textbf{(D) }10</math>. |
Solution by ngeorge | Solution by ngeorge | ||
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=B|before=-|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 10:45, 21 February 2016
Problem
What is the value of when ?
Solution
Factorizing the numerator, then becomes which is equal to which is .
Solution by ngeorge
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by - |
Followed by Problem 2 | |
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All AMC 10 Problems and Solutions |
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