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

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== Problem ==
 
== Problem ==
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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?
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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?
  
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== Solution ==
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==Solutions==
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===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.
  
 
Thus, <math>\dfrac{3k}{5k + 20} = \dfrac{29}{50}</math>, solving gives <math>k = 116</math>. Thus, the number of people that like to dance is <math>2k + 20 = \boxed{252}</math>.
 
Thus, <math>\dfrac{3k}{5k + 20} = \dfrac{29}{50}</math>, solving gives <math>k = 116</math>. Thus, the number of people that like to dance is <math>2k + 20 = \boxed{252}</math>.
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 +
===Solution 2===
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Let the number of girls be <math>g</math>. Let the number of total people originally be <math>t</math>.
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 +
We know that <math>\frac{g}{t}=\frac{3}{5}</math> from the problem.
 +
 +
We also know that <math>\frac{g}{t+20}=\frac{29}{50}</math> from the problem.
 +
 +
We now have a system and we can solve.
 +
 +
The first equation becomes:
 +
 +
<math>3t=5g</math>.
 +
 +
The second equation becomes:
 +
 +
<math>50g=29t+580</math>
 +
 +
Now we can sub in <math>30t=50g</math> by multiplying the first equation by <math>10</math>. We can plug this into our second equation.
 +
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<math>30t=29t+580</math>
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<math>t=580</math>
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We know that there were originally <math>580</math> people. Of those, <math>\frac{2}{5}*580=232</math> like to dance.
 +
 +
We also know that with these people, <math>20</math> boys joined, all of whom like to dance. We just simply need to add <math>20</math> to get <math>232+20=\boxed{252}</math>
 +
 +
==Solution 3==
 +
Let <math>p</math> denote the total number of people at the party. Then, because we know the proportions of boys to <math>p</math> both before and after 20 boys arrived, we can create the following equation:
 +
<cmath>0.4p+20 = 0.42(p+20)</cmath>
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Solving for p gives us <math>p=580</math>, so the solution is <math>0.4p+20 = \boxed{252}</math>
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 +
==Solution 4 (Cheese)==
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Assume all the boys like to dance and none of the girls like to dance. We then proceed like the previous solutions.
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~Arcticturn
  
 
== See also ==
 
== See also ==
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{{AIME box|year=2008|n=I|before=First question|num-a=2}}
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{{AIME box|year=2008|n=I|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:23, 19 December 2022

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?

Solutions

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 $232+20=\boxed{252}$

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 $0.4p+20 = \boxed{252}$

Solution 4 (Cheese)

Assume all the boys like to dance and none of the girls like to dance. We then proceed like the previous solutions.

~Arcticturn

See also

2008 AIME I (Problems • Answer Key • Resources)
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

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