Difference between revisions of "2021 AMC 10A Problems/Problem 2"
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− | ==Problem | + | ==Problem== |
Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have? | Portia's high school has <math>3</math> times as many students as Lara's high school. The two high schools have a total of <math>2600</math> students. How many students does Portia's high school have? | ||
<math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math> | <math>\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050</math> | ||
− | ==Solution== | + | ==Solution 1 (Two Variables)== |
− | The following system of equations can be formed with <math> | + | The following system of equations can be formed with <math>P</math> representing the number of students in Portia's high school and <math>L</math> representing the number of students in Lara's high school: |
− | <cmath> | + | <cmath>\begin{align*} |
− | + | P&=3L, \\ | |
− | Substituting <math> | + | P+L&=2600. |
+ | \end{align*}</cmath> | ||
+ | Substituting <math>P=3L</math> gives <math>4L=2600.</math> Solving for <math>L</math> gives <math>L=650.</math> Since we need to find <math>P,</math> we multiply <math>650</math> by <math>3</math> to get <math>P=\boxed{\textbf{(C)} ~1950}.</math> | ||
− | + | ~happykeeper (Solution) | |
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
==Solution 2 (One Variable)== | ==Solution 2 (One Variable)== | ||
− | Suppose Lara's high school has <math>x</math> students | + | Suppose Lara's high school has <math>x</math> students, so Portia's high school has <math>3x</math> students. We have <math>x+3x=2600,</math> or <math>4x=2600.</math> The answer is <cmath>3x=2600\cdot\frac 34=650\cdot3=\boxed{\textbf{(C)} ~1950}.</cmath> |
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Arithmetic)== | ||
+ | Clearly, <math>2600</math> is <math>4</math> times the number of students in Lara's high school. Therefore, Lara's high school has <math>2600\div4=650</math> students, and Portia's high school has <math>650\cdot3=\boxed{\textbf{(C)} ~1950}</math> students. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution | + | ==Solution 4 (Observations)== |
− | The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)} | + | The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)},\textbf{(D)},</math> and <math>\textbf{(E)}.</math> Since <math>\textbf{(A)}</math> is too small (as it is clear that <math>600+\frac{600}{3}<2600</math>), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math> |
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solutions== | ||
+ | ===Video Solution 1 (Very Fast & Simple)=== | ||
+ | https://youtu.be/DOtysU-a1B4 | ||
+ | |||
+ | ~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== | + | ===Video Solution 4 by OmegaLearn=== |
https://youtu.be/xXx0iP1tn8k | https://youtu.be/xXx0iP1tn8k | ||
− | + | ~pi_is_3.14 | |
+ | |||
+ | ===Video Solution 5=== | ||
+ | https://youtu.be/GwwDQYqptlQ | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ===Video Solution 6=== | ||
+ | https://youtu.be/50CThrk3RcM?t=66 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ===Video Solution 7 (Problems 1-3)=== | ||
+ | https://youtu.be/CupJpUzKPB0 | ||
+ | |||
+ | ~MathWithPi | ||
+ | |||
+ | ===Video Solution 8=== | ||
+ | https://youtu.be/slVBYmcDMOI | ||
+ | |||
+ | ~The Learning Royal | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2021|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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
~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.