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

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== Solution ==
 
== Solution ==
<math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\
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Sum the fractions over the common denominator: <math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45</math>
10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>
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Now the answer is just some arithmetic: <math>10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>
  
 
==See Also==
 
==See Also==

Revision as of 23:03, 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

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

See Also

2014 AMC 10A (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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