Difference between revisions of "2014 AMC 10A Problems/Problem 1"

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<math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\
 
<math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\
 
10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>
 
10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>
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==See Also==
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{{AMC10 box|year=2013|ab=B|before=First Problem|num-a=2}}
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{{MAA Notice}}

Revision as of 22:08, 6 February 2014

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

$\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\ 10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2$

See Also

2013 AMC 10B (ProblemsAnswer KeyResources)
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 10 Problems and Solutions

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