What is the value of (2^0-1+5^2+0)^-1 × 5?

2015 AMC 10A Problem 1 — Solution

2015 AMC 10A Problem 1

Attempt Problem 1 of the 2015 AMC 10A 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 10A solutions, or check the answer key.

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

$-125$

$-120$

$\dfrac{1}{5}$

$\dfrac{5}{24}$

$25$

Answer: C

Concepts:order of operations exponent

Difficulty rating: 560

Solution:

We can evaluate it as follows. $\begin{aligned} &(2^0-1+5^2+0)^{-1} \times 5 \\&=\dfrac{1}{1 - 1 + 25} \times 5 \\ &= \dfrac{5}{25}\\ &= \dfrac{1}{5} \end{aligned}$

Thus, C is the correct answer.

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