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

(Created page with "==Problem== Some boys and girls are having a car wash to raise money for a class trip to China. Initially <math>40\%</math> of the group are girls. Shortly thereafter two girls ...")
 
Line 1: Line 1:
 +
{{duplicate|[[2007 AMC 12B Problems|2007 AMC 12B #10]] and [[2007 AMC 10B Problems|2007 AMC 10B #14]]}}
 +
 
==Problem==
 
==Problem==
  
Line 15: Line 17:
  
 
==See Also==
 
==See Also==
 +
 +
{{AMC12 box|year=2007|ab=B|num-b=9|num-a=11}}
  
 
{{AMC10 box|year=2007|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2007|ab=B|num-b=13|num-a=15}}

Revision as of 15:24, 5 June 2011

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 $0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}$

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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