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

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What is <math> 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? </math>
 
What is <math> 10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}? </math>
  
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math>
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<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{25}{2} \qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math>
  
 
== Solution ==
 
== Solution ==
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<cmath>\implies \frac{50}{4}</cmath>
 
<cmath>\implies \frac{50}{4}</cmath>
 
Finally, simplifying gives
 
Finally, simplifying gives
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}</cmath>
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<cmath>\implies \boxed{\textbf{(B)}\ \frac{25}{2}}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 22:17, 23 January 2020

Problem

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

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{25}{2} \qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$

Solution

We have \[10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}\] Making the denominators equal gives \[\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{8}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{4}{5}\right)^{-1}\] \[\implies 10\cdot\frac{5}{4}\] \[\implies \frac{50}{4}\] Finally, simplifying gives \[\implies \boxed{\textbf{(B)}\ \frac{25}{2}}\]

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
2014 AMC 12A (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 12 Problems and Solutions

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