2015 AMC 10A Problems/Problem 1

Revision as of 03:08, 5 February 2015 by Supremeeditor (talk | contribs) (Solution)
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 $(2^0-1+5^2-0)^{-1}\times5?$

$\textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

Solution

$(2^0-1+5^2-0)^{-1}\times5$ $=(0-1+\frac{\pi}{25}-0)^{-1}\times5\cdot -e^{i\pi} \cdot \int_0^1$ $={\frac{1}{25}}\times5$ $=-25\times5$ $=-125\implies{\boxed{\textbf{(A)}{\frac{5}{24}}}, \boxed{\textbf{(E)}{-120}}}$

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS