Difference between revisions of "2008 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
| + | ==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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | Let the number of girls be <math>g</math>. Let the number of total people originally be <math>t</math>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | <math>30t=29t+580</math> | ||
| + | |||
| + | <math>t=580</math> | ||
| + | |||
| + | 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> | ||
== See also == | == See also == | ||
Revision as of 20:05, 20 November 2010
Problem
Of the students attending a school party, 
of the students are girls, and 
of the students like to dance. After these students are joined by 
more boy students, all of whom like to dance, the party is now 
girls. How many students now at the party like to dance?
Solution
Solution 1
Say that there were 
girls and 
boys at the party originally. like to dance. Then, there are girls and 
boys, and like to dance.
Thus, 
, solving gives 
. Thus, the number of people that like to dance is 
.
Solution 2
Let the number of girls be 
. Let the number of total people originally be 
.
We know that 
from the problem.
We also know that 
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 
by multiplying the first equation by 
. We can plug this into our second equation.
We know that there were originally 
people. Of those, 
like to dance.
We also know that with these people, boys joined, all of whom like to dance. We just simply need to add to get 
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 | ||
