Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math>, and <math>\textbf{(E)}</math>. Since <math>\textbf{(A)}</math> is too small (as <math>600+600/3<2600</math> is clearly true), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math> | The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math>, and <math>\textbf{(E)}</math>. Since <math>\textbf{(A)}</math> is too small (as <math>600+600/3<2600</math> is clearly true), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math> | ||
Revision as of 06:01, 14 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.
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
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.