Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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Latest revision as of 09:06, 11 July 2024
Contents
Problem
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 1 (Two Variables)
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:
Substituting 
gives 
Solving for gives 
Since we need to find 
we multiply 
by to get 
~happykeeper (Solution)
~MRENTHUSIASM (Reformatting)
Solution 2 (One Variable)
Suppose Lara's high school has 
students, so Portia's high school has 
students. We have 
or 
The answer is ![\[3x=2600\cdot\frac 34=650\cdot3=\boxed{\textbf{(C)} ~1950}.\]](http://latex.artofproblemsolving.com/5/2/7/52729c61ad5ca8860187c085e632e263b615531b.png)
~MRENTHUSIASM
Solution 3 (Arithmetic)
Clearly, is 
times the number of students in Lara's high school. Therefore, Lara's high school has 
students, and Portia's high school has 
students.
~MRENTHUSIASM
Solution 4 (Observations)
The number of students in Portia's high school must be a multiple of 
This eliminates 
and 
Since 
is too small (as it is clear that 
), we are left with 
~MRENTHUSIASM
Video Solutions
Video Solution 1 (Very Fast & Simple)
~Education, the Study of Everything
Video Solution 2 (Setting Variables)
https://youtu.be/qNf6SiIpIsk?t=119
~ThePuzzlr
Video Solution 3 (Solving by Equation)
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1
~North America Math Contest Go Go Go
Video Solution 4 by OmegaLearn
~pi_is_3.14
Video Solution 5
~savannahsolver
Video Solution 6
https://youtu.be/50CThrk3RcM?t=66
~IceMatrix
Video Solution 7 (Problems 1-3)
~MathWithPi
Video Solution 8
~The Learning Royal
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. 
