Difference between revisions of "2008 AIME I Problems/Problem 1"
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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> | 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 = 0.48(p+20)</cmath> | ||
+ | Solving for p gives us <math>p=580</math>, so the solution is <math>0.4p+20 = \boxed{252}</math> | ||
== See also == | == See also == |
Revision as of 14:10, 8 June 2018
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?
Solutions
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
Solution 3
Let denote the total number of people at the party. Then, because we know the proportions of boys to both before and after 20 boys arrived, we can create the following equation: Solving for p gives us , so the solution is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.