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

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==Problem 14==
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==Problem==
  
 
An Annville Junior High School, <math>30\%</math> of the students in the Math Club are in the Science Club, and <math>80\%</math> 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?
 
An Annville Junior High School, <math>30\%</math> of the students in the Math Club are in the Science Club, and <math>80\%</math> 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?
  
 
<math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40</math>
 
<math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 11:50, 23 December 2012

Problem

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