Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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+ | ==Video Solution 6== | ||
+ | https://youtu.be/50CThrk3RcM?t=66 | ||
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+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
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Revision as of 02:38, 19 February 2021
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
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
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
~MRENTHUSIASM
Solution 3 (Arithmetics)
Clearly, students is
times as many students as Lara's high school. Therefore, Lara's high school has
students, and Portia's high school has
students.
~MRENTHUSIASM
Solution 4 (Answer Choices)
Solution 4.1 (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
Solution 4.2 (Plug in the Answer Choices)
For we have
So,
is incorrect.
For we have
So,
is incorrect.
For we have
So,
is correct. For completeness, we will check choices
and
For we have
So,
is incorrect.
For we have
So,
is incorrect.
~MRENTHUSIASM
Video Solution #1(Setting Variables)
https://youtu.be/qNf6SiIpIsk?t=119 ~ThePuzzlr
Video Solution #2(Solving by equation)
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1 ~North America Math Contest Go Go Go
Video Solution
- pi_is_3.14
Video Solution (Simple)
~ Education, the Study of Everything
Video Solution 5
~savannahsolver
Video Solution 6
https://youtu.be/50CThrk3RcM?t=66
~IceMatrix
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.