Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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Revision as of 09:58, 12 February 2021
Contents
Problem 2
Portia's high school has 
times as many students as Lara's high school. The two high schools have a total of 
students. How many students does Portia's high school have?
Solution
The following system of equations can be formed with 
representing the number of students in Portia's high school and 
representing the number of students in Lara's high school.
![\[p=3q\]](http://latex.artofproblemsolving.com/7/6/a/76a779b71a5323a7719301624257488ce75dbd34.png)
![\[p+q=2600\]](http://latex.artofproblemsolving.com/a/a/a/aaac14ec573269e4ec0daaee0018ce402ed1d98a.png)
Substituting with 
we get 
. Solving for 
, we get 
. Since we need to find we multiply 
by 3 to get 
, which is 
-happykeeper
Solution 2 (One Variable)
Suppose Lara's high school has 
students. It follows that Portia's high school has 
students. We know that 
or 
Our answer is ![\[3x=2600\left(\frac 34\right)=650(3)=\boxed{\textbf{(C)} ~1950}.\]](http://latex.artofproblemsolving.com/a/8/d/a8d4e4eeb5655a2cf15418c46bed0cf937e4c6c3.png)
~MRENTHUSIASM
Solution 3 (Quick Inspection)
The number of students in Portia's high school must be a multiple of 
This eliminates 
, 
, and 
. Since 
is too small (as 
is clearly true), we are left with 
~MRENTHUSIASM
Video Solution #1(Setting Variables)
https://youtu.be/qNf6SiIpIsk?t=119 ~ThePuzzlr
Video Solution
- pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
