Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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==Solution 2 (One Variable)== | ==Solution 2 (One Variable)== | ||
Suppose Lara's high school has <math>x</math> students. It follows that Portia's high school has <math>3x</math> students. We know that <math>x+3x=2600,</math> or <math>4x=2600.</math> Our answer is <cmath>3x=2600\left(\frac 34\right)=650(3)=\boxed{\text{(C) }1950}.</cmath> | Suppose Lara's high school has <math>x</math> students. It follows that Portia's high school has <math>3x</math> students. We know that <math>x+3x=2600,</math> or <math>4x=2600.</math> Our answer is <cmath>3x=2600\left(\frac 34\right)=650(3)=\boxed{\text{(C) }1950}.</cmath> | ||
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| + | ~MRENTHUSIASM | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 21:22, 11 February 2021
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 ![\[3x=2600\left(\frac 34\right)=650(3)=\boxed{\text{(C) }1950}.\]](http://latex.artofproblemsolving.com/6/3/1/6315a780f5deebcd04a31143ed2560b07a26d02c.png)
~MRENTHUSIASM
Video Solution
- pi_is_3.14
