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2021 AMC 10A Problems/Problem 2: Difference between revisions

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<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>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>
<cmath>\begin{align*}
<cmath>p+q=2600</cmath>
P&=3L, \\
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>
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
~happykeeper (Solution)
 
~MRENTHUSIASM (Reformatting)


==Solution 2 (One Variable)==
==Solution 2 (One Variable)==
Suppose Lara's high school has <math>x</math> students. It follows that Portia's high school has <math>3x</math> students. We know that <math>x+3x=2600,</math> or <math>4x=2600.</math> Our answer is <cmath>3x=2600\left(\frac 34\right)=650(3)=\boxed{\textbf{(C)} ~1950}.</cmath>
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
~MRENTHUSIASM


==Solution 3 (Arithmetics)==
==Solution 3 (Arithmetic)==
Clearly, <math>2600</math> students is <math>4</math> times as many students as 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.
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 4 (Answer Choices)==
==Solution 4 (Observations)==
===Solution 4.1 (Quick Inspection)===
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>
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 <math>600+600/3<2600</math> is clearly true), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math>


~MRENTHUSIASM
~MRENTHUSIASM


===Solution 4.2 (Plug in the Answer Choices)===
==Video Solutions==
For <math>\textbf{(A)},</math> we have <math>600+\frac{600}{3}=800\neq2600.</math> So, <math>\textbf{(A)}</math> is incorrect.
===Video Solution 1 (Very Fast & Simple)===
https://youtu.be/DOtysU-a1B4


For <math>\textbf{(B)},</math> we have <math>650+\frac{650}{3}=866\frac{2}{3}\neq2600.</math> So, <math>\textbf{(B)}</math> is incorrect.
~Education, the Study of Everything


For <math>\textbf{(C)},</math> we have <math>1950+\frac{1950}{3}=2600.</math> So, <math>\boxed{\textbf{(C)} ~1950}</math> is correct. For completeness, we will check choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math>
===Video Solution 3 (Solving by Equation)===


For <math>\textbf{(D)},</math> we have <math>2000+\frac{2000}{3}=2666\frac{2}{3}\neq2600.</math> So, <math>\textbf{(D)}</math> is incorrect.
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1


For <math>\textbf{(E)},</math> we have <math>2050+\frac{2050}{3}=2733\frac{1}{3}\neq2600.</math> So, <math>\textbf{(E)}</math> is incorrect.
~North America Math Contest Go Go Go


~MRENTHUSIASM
===Video Solution 4 by OmegaLearn===
https://youtu.be/xXx0iP1tn8k


==Video Solution #1(Setting Variables) ==
~pi_is_3.14


https://youtu.be/qNf6SiIpIsk?t=119
===Video Solution 5===
~ThePuzzlr
https://youtu.be/GwwDQYqptlQ


==Video Solution #2(Solving by equation) ==
~savannahsolver


https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1
===Video Solution 6===
~North America Math Contest Go Go Go
https://youtu.be/50CThrk3RcM?t=66
 
==Video Solution==
https://youtu.be/xXx0iP1tn8k


- pi_is_3.14
~IceMatrix


==Video Solution (Simple)==
===Video Solution 7 (Problems 1-3)===
https://youtu.be/DOtysU-a1B4
https://youtu.be/CupJpUzKPB0


~Education, the Study of Everything
~MathWithPi


==Video Solution 5==
===Video Solution 8===
https://youtu.be/GwwDQYqptlQ
https://youtu.be/slVBYmcDMOI


~savannahsolver
~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 07:26, 15 July 2025

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 (Two Variables)

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: \begin{align*} P&=3L, \\ P+L&=2600. \end{align*} Substituting $P=3L$ gives $4L=2600.$ Solving for $L$ gives $L=650.$ Since we need to find $P,$ we multiply $650$ by $3$ to get $P=\boxed{\textbf{(C)} ~1950}.$

~happykeeper (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2 (One Variable)

Suppose Lara's high school has $x$ students, so Portia's high school has $3x$ students. We have $x+3x=2600,$ or $4x=2600.$ The answer is \[3x=2600\cdot\frac 34=650\cdot3=\boxed{\textbf{(C)} ~1950}.\]

~MRENTHUSIASM

Solution 3 (Arithmetic)

Clearly, $2600$ is $4$ times the number of students in 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 (Observations)

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 it is clear that $600+\frac{600}{3}<2600$), we are left with $\boxed{\textbf{(C)} ~1950}.$

~MRENTHUSIASM

Video Solutions

Video Solution 1 (Very Fast & Simple)

https://youtu.be/DOtysU-a1B4

~Education, the Study of Everything

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

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

2021 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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