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2007 AMC 10B Problems/Problem 14: Difference between revisions

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<math>\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12</math>
<math>\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12</math>


==Solution==
==Solution 1==


If we let <math>p</math> be the number of people initially in the group, then <math>0.4p</math> is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still <math>p</math>, but the number of girls is <math>0.4p-2</math>. Since only <math>30\%</math> of the group are girls,
If we let <math>p</math> be the number of people initially in the group, then <math>0.4p</math> is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still <math>p</math>, but the number of girls is <math>0.4p-2</math>. Since only <math>30\%</math> of the group are girls,
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The number of girls initially in the group is <math>0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}</math>
The number of girls initially in the group is <math>0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}</math>


==Alternate Solution==
==Solution 2==


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


==Yet Another Alternate Solution==
==Solution 3==


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>.

Latest revision as of 22:19, 29 September 2023

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 1

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

Solution 2

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

Solution 3

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 $g = \boxed{ 8\ \mathrm{(C)}}$.

~mobius247

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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