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= \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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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>\boxed{\textbf{(D) }10}</math>. | 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>\boxed{\textbf{(D) }10}</math>. | ||
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| + | ==Solution 2== | ||
| + | Substituting <math>\frac{1}{2}</math> for <math>a</math> in <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> gives us <math>\boxed{\textbf{(D) }10}</math>. | ||
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| + | Remember, <math>x^{-y}=\frac{1}{x^y}</math>! | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/2erUXM5pD2g | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | |||
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| + | ==Video Solution== | ||
| + | https://youtu.be/1IZ3oj1iGf0 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|before=-|num-a=2}} | {{AMC10 box|year=2016|ab=B|before=-|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category: Introductory Algebra Problems]] | ||
Latest revision as of 16:54, 18 October 2025
Contents
Problem
What is the value of 
when 
?
Solution
Factorizing the numerator, 
then becomes 
which is equal to 
which is 
.
Solution 2
Substituting 
for 
in gives us .
Remember, 
!
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
