Difference between revisions of "2008 AIME I Problems/Problem 1"
m (→Problem) |
Arcticturn (talk | contribs) (→Solution 3) |
||
(11 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Of the students attending a school party, <math>60</math> | + | 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 == | + | ==Solutions== |
+ | ===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> | ||
+ | |||
+ | ==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> | ||
+ | Solving for p gives us <math>p=580</math>, so the solution is <math>0.4p+20 = \boxed{252}</math> | ||
+ | |||
+ | ==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 == | == See also == | ||
− | {{AIME box|year=2008|n=I|before=First | + | {{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 22:23, 19 December 2022
Contents
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.