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 | + | 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> | + | |
| + | 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 
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$ (Error compiling LaTeX. Unknown error_msg)
Solution
Sum the fractions over the common denominator: 
Now the answer is just some arithmetic: 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
