Difference between revisions of "2015 AMC 10A Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | <math>(2^0-1+5^2-0)^{-1}\times5 = (1-1+25-0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \boxed{\textbf{(C) \, \frac{1}{5}}</math>. | + | <math>(2^0-1+5^2-0)^{-1}\times5 = (1-1+25-0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \boxed{\textbf{(C) } \, \frac{1}{5}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 10:57, 5 February 2015
- The following problem is from both the 2015 AMC 12A #1 and 2015 AMC 10A #1, so both problems redirect to this page.
Problem
What is the value of 
$\textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25$ (Error compiling LaTeX. Unknown error_msg)
Solution
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See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
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 | ||
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
