Difference between revisions of "2000 AMC 10 Problems/Problem 8"

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==Problem==
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==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>.
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<math>\frac{2}{5}f=\frac{4}{5}s</math>
<math>f=2s</math>.
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<math>f=2s</math>
  
 
There are twice as many freshman as sophomores.
 
There are twice as many freshman as sophomores.
D.
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<math>\boxed{\text{D}}</math>
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==See Also==
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{{AMC10 box|year=2000|num-b=7|num-a=9}}

Revision as of 19:35, 8 January 2009

Problem

Solution

Let $f$ be the number of freshman and s be the number of sophomores.

$\frac{2}{5}f=\frac{4}{5}s$

$f=2s$

There are twice as many freshman as sophomores. $\boxed{\text{D}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
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
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