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2014 AMC 12A Problem 1

Problem 1 of 25EasierArithmetic

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

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Solution

The sum inside the parentheses is 12+15+110=5+2+110=45.\dfrac12+\dfrac15+\dfrac{1}{10}=\dfrac{5+2+1}{10}=\dfrac{4}{5}. Its reciprocal is 54,\dfrac54, so the expression equals 10⋅54=252.10\cdot\dfrac54=\dfrac{25}{2}. Thus, the correct answer is C.
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