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

Latest revision as of 22:08, 26 June 2023

The following problem is from both the 2015 AMC 12A #1 and 2015 AMC 10A #1, so both problems redirect to this page.

What is the value of $(2^0-1+5^2-0)^{-1}\times5?$

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

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

https://youtu.be/yH4KLfe9p88

~Education, the Study of Everything

~savannahsolver

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #1]] and [[2015 AMC 10A Problems|2015 AMC 10A #1]]}}
+==Problem==
+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>
 ==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>.
+==Video Solution (CREATIVE THINKING)==
+ https://youtu.be/yH4KLfe9p88
+~Education, the Study of Everything
+==Video Solution==
+https://youtu.be/UdENmDoPGHU
+~savannahsolver
-Simply evaluating gives <math>\boxed{\textbf{(C)}frac15}</math>.
+==See Also==
+{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
+{{AMC12 box|year=2015|ab=A|before=First Problem|num-a=2}}
+{{MAA Notice}}