2015 AMC 12A Problem 1

Attempt Problem 1 of the 2015 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12A solutions, or check the answer key.

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1.

What is the value of (201+52+0)1×5?(2^0-1+5^2+0)^{-1}\times 5?

125-125

120-120

15\dfrac{1}{5}

524\dfrac{5}{24}

2525

Answer: C
Concepts:order of operationsexponent
Difficulty rating: 890
Solution:

Inside the parentheses, 201+52+0=11+25+0=25. \begin{aligned} &2^0-1+5^2+0 \\ &\quad = 1-1+25+0 = 25. \end{aligned}

Then (25)1×5=525=15.(25)^{-1}\times 5 = \dfrac{5}{25} = \dfrac{1}{5}.

Thus, the correct answer is C.

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