Difference between revisions of "2015 AMC 10A Problems/Problem 1"

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

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

Solution

$(2^0-1+5^2-0)^{-1}\times5 = (1-1+25-0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \boxed{\textbf{(C) \, \frac{1}{5}}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.).

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

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