Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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<cmath>\implies \frac{50}{4}</cmath> | <cmath>\implies \frac{50}{4}</cmath> | ||
<cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath> | <cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2014|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Revision as of 10:15, 9 February 2014
Solution
We have ![\[10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/1/2/e/12e21260e737c4ecc2bf7b4116b1f1e7bcde2b37.png)
![\[\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/8/5/8/858522583ebacde2c4e851b8fd64c093086a668a.png)
![\[\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/b/9/c/b9c7e18416c20508fb4b4e0822b1c6183699a746.png)
![\[\implies 10\cdot\left(\frac{8}{10}\right)^{-1}\]](http://latex.artofproblemsolving.com/2/8/3/2832b440e25ccb94bfdd37a6c04898ff0c7544ca.png)
![\[\implies 10\cdot\left(\frac{4}{5}\right)^{-1}\]](http://latex.artofproblemsolving.com/6/d/3/6d309b6b45b8f4e35d741ae32ea81647183f0b5e.png)
![\[\implies 10\cdot\frac{5}{4}\]](http://latex.artofproblemsolving.com/8/6/1/8612999285ee03d02effd49381ab5a5a3aea86a6.png)
![\[\implies \frac{50}{4}\]](http://latex.artofproblemsolving.com/0/1/0/0100cd85a3f1692e5c483ab0559633686ad33cdd.png)
![\[\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare\]](http://latex.artofproblemsolving.com/7/2/b/72b9f35167c38595fb942bf6a1efdc21bbb27489.png)
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
