Difference between revisions of "1998 AJHSME Problems/Problem 14"

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==Solution==
 
==Solution==
  
If <math>80\%</math> of the people in the science club of <math>15</math> people are in the Math Club, <math>\frac{4}{5}\times15=12</math> people are in the Math Club who is in the Science Club.
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If <math>80\%</math> of the people in the science club of <math>15</math> people are in the Math Club, <math>\frac{4}{5}\times15=12</math> people are in the both the Math Club and the Science Club.
  
<math>12</math> is also <math>30\%</math> compared to how many people are in the Math Club.
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These <math>12</math> people are also <math>30\%</math> of the Math Club.
  
<math>12=30\%</math>
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Setting up a proportion:
  
<math>4=10\%</math>
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<math>\frac{12}{30\%} = \frac{x}{100\%}</math>
  
<math>40=100\%</math>
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<math>12\cdot 1.00 =0.30\cdot x</math>
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<math>\frac{12}{0.3} = 40=x </math>
  
 
There are <math>40=\boxed{E}</math> people in the Math Club.
 
There are <math>40=\boxed{E}</math> people in the Math Club.
 
  
 
== See also ==
 
== See also ==

Revision as of 11:33, 31 July 2011

Problem 14

An Annville Junior High School, $30\%$ of the students in the Math Club are in the Science Club, and $80\%$ of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club?

$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40$


Solution

If $80\%$ of the people in the science club of $15$ people are in the Math Club, $\frac{4}{5}\times15=12$ people are in the both the Math Club and the Science Club.

These $12$ people are also $30\%$ of the Math Club.

Setting up a proportion:

$\frac{12}{30\%} = \frac{x}{100\%}$

$12\cdot 1.00 =0.30\cdot x$

$\frac{12}{0.3} = 40=x$

There are $40=\boxed{E}$ people in the Math Club.

See also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions