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Difference between revisions of "2007 AMC 10B Problems/Problem 14"

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p&=20\end{align*}</cmath>
 
p&=20\end{align*}</cmath>
 
The number of girls is <math>0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}</math>
 
The number of girls is <math>0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}</math>
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==Alternate Solution==
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There are the same number of total people before and after, but the number of girls has dropped by two and <math>10\%</math>.  <math>\frac{2}{0.1}=20</math>, and <math>40\%\cdot20=8</math>, so the answer is <math>\mathrm{(C)}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 12:48, 21 January 2015

The following problem is from both the 2007 AMC 12B #10 and 2007 AMC 10B #14, so both problems redirect to this page.

Problem

Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?

$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$

Solution

If we let $p$ be the number of people initially in the group, the $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p,$ but the number of girls is $0.4p-2$. Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=3p\\ p&=20\end{align*} The number of girls is $0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}$

Alternate Solution

There are the same number of total people before and after, but the number of girls has dropped by two and $10\%$. $\frac{2}{0.1}=20$, and $40\%\cdot20=8$, so the answer is $\mathrm{(C)}$.

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions
2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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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