2008 AIME I Problems/Problem 1


Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?


Solution 1

Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.

Thus, $\dfrac{3k}{5k + 20} = \dfrac{29}{50}$, solving gives $k = 116$. Thus, the number of people that like to dance is $2k + 20 = \boxed{252}$.

Solution 2

Let the number of girls be $g$. Let the number of total people originally be $t$.

We know that $\frac{g}{t}=\frac{3}{5}$ from the problem.

We also know that $\frac{g}{t+20}=\frac{29}{50}$ from the problem.

We now have a system and we can solve.

The first equation becomes:


The second equation becomes:


Now we can sub in $30t=50g$ by multiplying the first equation by $10$. We can plug this into our second equation.



We know that there were originally $580$ people. Of those, $\frac{2}{5}*580=232$ like to dance.

We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=\boxed{252}$

Solution 3

Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$, so the solution is $0.4p+20 = \boxed{252}$

See also

2008 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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

Invalid username
Login to AoPS