Difference between revisions of "2000 AMC 10 Problems/Problem 8"
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| + | ==Problem== | ||
| + | |||
| + | ==Solution== | ||
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Let <math>f</math> be the number of freshman and s be the number of sophomores. | Let <math>f</math> be the number of freshman and s be the number of sophomores. | ||
| − | <math>\frac{2}{5}f=\frac{4}{5}s</math> | + | <math>\frac{2}{5}f=\frac{4}{5}s</math> |
| − | <math>f=2s</math> | + | |
| + | <math>f=2s</math> | ||
There are twice as many freshman as sophomores. | There are twice as many freshman as sophomores. | ||
| − | D | + | <math>\boxed{\text{D}}</math> |
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2000|num-b=7|num-a=9}} | ||
Revision as of 18:35, 8 January 2009
Problem
Solution
Let 
be the number of freshman and s be the number of sophomores.
There are twice as many freshman as sophomores.
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
