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

Revision as of 13:04, 4 June 2021

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, then $0.4p$ 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 initially in the group 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 $\mathrm{(C)}$ .

Yet Another Alternate Solution

Let $x$ be the number of people initially in the group and $g$ the number of girls. $\frac{2}{5}x = g$ , so $x = \frac{5}{2}g$ . Also, the problem states $\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \boxed{ 8\mathrm{(C)}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==Yet Another Alternate Solution==
-Let <math>x</math> be the number of people initially in the group and <math>g</math> the number of girls. <math>\frac{2}{5}x = g</math>, so <math>x = \frac{5}{2}g</math>. Also, the problem states <math>\frac{3}{10}x = g-2</math>. Substituting <math>x</math> in terms of <math>g</math> into the second equation yields that <math>g = \boxed{ 8\mathrm{(C) \ }}</math>.
+Let <math>x</math> be the number of people initially in the group and <math>g</math> the number of girls. <math>\frac{2}{5}x = g</math>, so <math>x = \frac{5}{2}g</math>. Also, the problem states <math>\frac{3}{10}x = g-2</math>. Substituting <math>x</math> in terms of <math>g</math> into the second equation yields that <math>g = \boxed{ 8\mathrm{(C)}}</math>.
 ==See Also==

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