2014 AMC 12A Problems/Problem 1

Problem

What is $10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?$

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$ (Error compiling LaTeX. Unknown error_msg)

Solution

Sum the fractions over the common denominator: $\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45$

Now the answer is just some arithmetic: $10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}= \boxed{\textbf{(C)}\ \dfrac{25}2}$

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 12A Problems/Problem 1

Problem

Solution

See Also