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 | |
| 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 AMC 10 Problems and Solutions | ||
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