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2014 AMC 10A Problem 1

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2014 AMC 10A Problem 1AMC 10 Step by Step
Name:                             Date: September 13, 2026
Fractions & Decimals · difficulty 1 of 5 · about 1 min · position P1-10

What is 10⋅(12+15+110)−1?10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?

  1. A)

    33

  2. B)

    88

  3. C)

    252\frac{25}{2}

  4. D)

    1703\frac{170}{3}

  5. E)

    170170

Answer

C

Key insight

Add the three fractions over the common denominator 10 to get 4/5, then the exponent -1 flips it to 5/4 before multiplying by 10.

Solution

Work inside the parentheses first, using denominator 1010:
12+15+110=510+210+110=810=45.\frac{1}{2} + \frac{1}{5} + \frac{1}{10} = \frac{5}{10} + \frac{2}{10} + \frac{1}{10} = \frac{8}{10} = \frac{4}{5}.
The exponent −1-1 takes the reciprocal, so the parenthesized quantity becomes 54\frac{5}{4}. Finally
10⋅54=504=252.10 \cdot \frac{5}{4} = \frac{50}{4} = \frac{25}{2}.

The answer is (C) 252\boxed{\textbf{(C)}\ \frac{25}{2}}.

Why this works

Order of operations: the sum inside the parentheses must be completed before the reciprocal is taken, because (p+q)−1(p+q)^{-1} is not p−1+q−1p^{-1} + q^{-1}. Whenever a negative exponent sits on a sum, simplify the sum to a single fraction first, then flip it.

The trap

Applying the -1 exponent to each fraction separately, adding 2 + 5 + 10 = 17, and choosing (E) 170.

Common mistakes

  • Applying the -1 exponent to each fraction separately, adding 2 + 5 + 10 = 17, and choosing (E) 170.
  • Forgetting to take the reciprocal at all and computing 10⋅45=810 \cdot \frac{4}{5} = 8, choice (B).

Techniques

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