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2015 AMC 10A Problem 1

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2015 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

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

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

Answer

Key insight

Inside the parentheses 2^0 = 1 cancels the -1, leaving 25; a negative-one exponent means reciprocal, so the value is 5/25.

Solution

Work inside the parentheses first. Since $2^0 = 1$ and $5^2 = 25$ ,
$2^0 - 1 + 5^2 + 0 = 1 - 1 + 25 + 0 = 25.$
The exponent $-1$ turns $25$ into its reciprocal $\frac{1}{25}$ . Multiplying by $5$ :
$\frac{1}{25} \times 5 = \frac{5}{25} = \frac{1}{5}.$

The answer is $\boxed{\textbf{(C)}\ \frac{1}{5}}$ .

Why this works

This is a pure order-of-operations check: evaluate exponents, then the sum inside the parentheses, then the outer exponent, then the multiplication. The two facts being tested are $a^0 = 1$ for $a \neq 0$ and $a^{-1} = \frac{1}{a}$ ; keep both straight and nothing else can go wrong.

The trap

Treating 2^0 as 0 (getting 24 inside, choice D) or applying the exponent -1 as a negative sign instead of a reciprocal.

Common mistakes

Treating $2^0$ as $0$ (getting $24$ inside, choice D) or applying the exponent $-1$ as a negative sign instead of a reciprocal.
Multiplying $25 \times 5 = 125$ and then attaching a minus sign, which is how choice (A) is built.

Techniques

Set up the equation/formula and compute; no special trick needed