Difference between revisions of "2012 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | The ratio <math>\frac{400 \text{ euros}}{500 \text{ dollars}}</math> can be simplified using | + | The ratio <math>\frac{400 \text{ euros}}{500 \text{ dollars}}</math> can be simplified using conversion factors:<cmath>\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04</cmath> which means the money is greater by <math>\boxed{ \textbf{(B)} \ 4 }</math> percent. |
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== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2012|ab=B|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 13:05, 5 July 2013
Problem
Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
Solution
The ratio 
can be simplified using conversion factors:![\[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\]](http://latex.artofproblemsolving.com/7/e/9/7e910f71f1672e328c980f1d1de03924380a02f8.png)
which means the money is greater by 
percent.
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
