Difference between revisions of "2016 AMC 10B Problems/Problem 1"
m (→Solution) |
(→Solution 2) |
||
(16 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
==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> | ||
+ | ==Solution== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/2erUXM5pD2g | ||
+ | ~Education, the Study of Everything | ||
− | |||
− | + | ==Video Solution== | |
+ | https://youtu.be/1IZ3oj1iGf0 | ||
+ | |||
+ | ~savannahsolver | ||
− | + | ==See Also== | |
+ | {{AMC10 box|year=2016|ab=B|before=-|num-a=2}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:39, 6 August 2024
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.