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

(See Also)
Line 1: Line 1:
{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #1]] and [[2014 AMC 10A Problems|2014 AMC 10A #1]]}}
 
 
==Problem ==
 
 
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>
 
 
 
== Solution ==
 
== Solution ==
Sum the fractions over the common denominator: <math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45</math>
 
 
Now the answer is just some arithmetic: <math>10\cdot\left(\dfrac45\right)^{-1}=10\cdot\dfrac{5}{4}=\boxed{\textbf{(C)}\ \dfrac{25}2}</math>
 
 
==See Also==
 
  
{{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}
+
We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath>
{{AMC12 box|year=2014|ab=A|before=First Problem|num-a=2}}
+
<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath>
{{MAA Notice}}
+
<cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath>
 +
<cmath>\implies 10\cdot\left(\frac{8}{10}\right)^{-1}</cmath>
 +
<cmath>\implies 10\cdot\left(\frac{4}{5}\right)^{-1}</cmath>
 +
<cmath>\implies 10\cdot\frac{5}{4}</cmath>
 +
<cmath>\implies \frac{50}{4}</cmath>
 +
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath>

Revision as of 10:11, 9 February 2014

Solution

We have \[10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{8}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{4}{5}\right)^{-1}\] \[\implies 10\cdot\frac{5}{4}\] \[\implies \frac{50}{4}\] \[\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare\]