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

(Solution)
(Solution)
(14 intermediate revisions by 6 users not shown)
Line 4: Line 4:
 
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math>
 
What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math>
  
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math>
+
<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math>
  
 
==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>=(1-1+\{25}-0)^{-1}\times5</math>
+
 
<math>={\frac{1}{25}}\times5</math>
+
==Video Solution==
<math>=-25\times5</math>
+
https://youtu.be/UdENmDoPGHU
<math>=-125\implies{\boxed{\textbf{(C)}\ -5}}</math>
+
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 09:49, 15 September 2022

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 $(2^0-1+5^2-0)^{-1}\times5?$

$\textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25$

Solution

$(2^0-1+5^2-0)^{-1}\times5 = (1-1+25-0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \boxed{\textbf{(C) } \, \frac{1}{5}}$.

Video Solution

https://youtu.be/UdENmDoPGHU

~savannahsolver

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