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Difference between revisions of "2021 AMC 10A Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
 
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.  
 
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>p=3q</cmath>
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<cmath>p=3l</cmath>
−
<cmath>p+q=2600</cmath>
+
<cmath>p+l=2600</cmath>
−
Substituting <math>p</math> with <math>3q</math> we get <math>4q=2600</math>. Solving for <math>q</math>, we get <math>q=650</math>. Since we need to find <math>p</math> we multiply <math>650</math> by 3 to get <math>p=1950</math>, which is <math>\boxed{\text{C}}</math>
+
Substituting <math>p</math> with <math>3l</math> we get <math>4l=2600</math>. Solving for <math>l</math>, we get <math>l=650</math>. Since we need to find <math>p</math> we multiply <math>650</math> by 3 to get <math>p=1950</math>, which is <math>\boxed{\text{C}}</math>
  
 
-happykeeper
 
-happykeeper

Revision as of 23:07, 21 February 2021

Problem

Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?

$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$

Solution 1

The following system of equations can be formed with $p$ representing the number of students in Portia's high school and $l$ representing the number of students in Lara's high school. \[p=3l\] \[p+l=2600\] Substituting $p$ with $3l$ we get $4l=2600$. Solving for $l$, we get $l=650$. Since we need to find $p$ we multiply $650$ by 3 to get $p=1950$, which is $\boxed{\text{C}}$

-happykeeper

Solution 2 (One Variable)

Suppose Lara's high school has $x$ students. It follows that Portia's high school has $3x$ students. We know that $x+3x=2600,$ or $4x=2600.$ Our answer is \[3x=2600\left(\frac 34\right)=650(3)=\boxed{\textbf{(C)} ~1950}.\]

~MRENTHUSIASM

Solution 3 (Arithmetics)

Clearly, $2600$ students is $4$ times as many students as Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{\textbf{(C)} ~1950}$ 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 $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}$. Since $\textbf{(A)}$ is too small (as $600+600/3<2600$ is clearly true), we are left with $\boxed{\textbf{(C)} ~1950}.$

~MRENTHUSIASM

Solution 4.2 (Plug in the Answer Choices)

For $\textbf{(A)},$ we have $600+\frac{600}{3}=800\neq2600.$ So, $\textbf{(A)}$ is incorrect.

For $\textbf{(B)},$ we have $650+\frac{650}{3}=866\frac{2}{3}\neq2600.$ So, $\textbf{(B)}$ is incorrect.

For $\textbf{(C)},$ we have $1950+\frac{1950}{3}=2600.$ So, $\boxed{\textbf{(C)} ~1950}$ is correct. For completeness, we will check choices $\textbf{(D)}$ and $\textbf{(E)}.$

For $\textbf{(D)},$ we have $2000+\frac{2000}{3}=2666\frac{2}{3}\neq2600.$ So, $\textbf{(D)}$ is incorrect.

For $\textbf{(E)},$ we have $2050+\frac{2050}{3}=2733\frac{1}{3}\neq2600.$ So, $\textbf{(E)}$ 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

https://youtu.be/xXx0iP1tn8k

- pi_is_3.14

Video Solution (Simple)

https://youtu.be/DOtysU-a1B4

~ Education, the Study of Everything

Video Solution 5

https://youtu.be/GwwDQYqptlQ

~savannahsolver

Video Solution 6

https://youtu.be/50CThrk3RcM?t=66

~IceMatrix

Video Solution (Problems 1-3)

https://youtu.be/CupJpUzKPB0

~MathWithPi

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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