Difference between revisions of "2007 AMC 10B Problems/Problem 14"
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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> | ||
| − | == | + | ==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>. | ||
| − | == | + | ==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 
of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then 
of the group are girls. How many girls were initially in the group?
Solution 1
If we let 
be the number of people initially in the group, then 
is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still , but the number of girls is 
. Since only of the group are girls,
The number of girls initially in the group is 
Solution 2
There are the same number of total people before and after, but the number of girls has dropped by two or 
of the total. 
, and 
, so the answer is 
.
Solution 3
Let 
be the number of people initially in the group and 
the number of girls. 
, so 
. Also, the problem states 
. Substituting in terms of into the second equation yields that 
.
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
