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

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==Solution==
 
==Solution==
  
Let <math>f</math> be the number of freshman and s be the number of sophomores.
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Let <math>f</math> be the number of freshman and <math>s</math> 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>

Revision as of 11:16, 24 January 2009

Problem

At Olympic High School, $\frac{2}{5}$ of the freshmen and $\frac{4}{5}$ of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?

$\mathrm{(A)}$ There are five times as many sophomores as freshmen.

$\mathrm{(B)}$ There are twice as many sophomores as freshmen.

$\mathrm{(C)}$ There are as many freshmen as sophomores.

$\mathrm{(D)}$ There are twice as many freshmen as sophomores.

$\mathrm{(E)}$ There are five times as many freshmen as sophomores.

Solution

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

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

$2f = 4s$

$f=2s$

There are twice as many freshmen 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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