Difference between revisions of "2008 AIME I Problems/Problem 1"

m (Solution 2)
m (Solution 1)
Line 2: Line 2:
 
Of the students attending a school party, <math>60\%</math> of the students are girls, and <math>40\%</math> of the students like to dance. After these students are joined by <math>20</math> more boy students, all of whom like to dance, the party is now <math>58\%</math> girls. How many students now at the party like to dance?
 
Of the students attending a school party, <math>60\%</math> of the students are girls, and <math>40\%</math> of the students like to dance. After these students are joined by <math>20</math> more boy students, all of whom like to dance, the party is now <math>58\%</math> girls. How many students now at the party like to dance?
  
 +
==Solution==
 
===Solution 1===
 
===Solution 1===
 
Say that there were <math>3k</math> girls and <math>2k</math> boys at the party originally. <math>2k</math> like to dance. Then, there are <math>3k</math> girls and <math>2k + 20</math> boys, and <math>2k + 20</math> like to dance.
 
Say that there were <math>3k</math> girls and <math>2k</math> boys at the party originally. <math>2k</math> like to dance. Then, there are <math>3k</math> girls and <math>2k + 20</math> boys, and <math>2k + 20</math> like to dance.

Revision as of 19:18, 5 January 2014

Problem

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

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:

$3t=5g$.

The second equation becomes:

$50g=29t+580$

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

$30t=29t+580$

$t=580$

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 $232+20=\boxed{252}$

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