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

(Solution)
(Solution)
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We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath>
 
We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath>
 +
Making the denominators equal gives
 
<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath>
 
<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath>
 
<cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath>
 
<cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath>
Line 13: Line 14:
 
<cmath>\implies 10\cdot\frac{5}{4}</cmath>
 
<cmath>\implies 10\cdot\frac{5}{4}</cmath>
 
<cmath>\implies \frac{50}{4}</cmath>
 
<cmath>\implies \frac{50}{4}</cmath>
 +
Finally, simplifying gives
 
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath>
 
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath>
  

Revision as of 11:19, 9 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

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 \[\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare\]

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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