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

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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 1==
+
==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=3l</cmath>
 
<cmath>p=3l</cmath>
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==Solution 3 (Arithmetic)==
 
==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.
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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
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
==Video Solution (Very fast & Simple)==
+
==Video Solution 1 (Very Fast & Simple)==
 
https://youtu.be/DOtysU-a1B4
 
https://youtu.be/DOtysU-a1B4
  
 
~ Education, the Study of Everything
 
~ Education, the Study of Everything
  
 
+
==Video Solution 2 (Setting Variables) ==
==Video Solution #1(Setting Variables) ==
 
  
 
https://youtu.be/qNf6SiIpIsk?t=119
 
https://youtu.be/qNf6SiIpIsk?t=119
 
~ThePuzzlr
 
~ThePuzzlr
  
==Video Solution #2(Solving by equation) ==
+
==Video Solution 3 (Solving by Equation) ==
  
 
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1
 
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1
 
~North America Math Contest Go Go Go
 
~North America Math Contest Go Go Go
  
==Video Solution==
+
==Video Solution 4==
 
https://youtu.be/xXx0iP1tn8k
 
https://youtu.be/xXx0iP1tn8k
  
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~IceMatrix
 
~IceMatrix
  
==Video Solution (Problems 1-3)==
+
==Video Solution 7 (Problems 1-3)==
 
https://youtu.be/CupJpUzKPB0
 
https://youtu.be/CupJpUzKPB0
  

Latest revision as of 19:58, 26 May 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 (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. \[p=3l\] \[p+l=2600\] 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, 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 (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 it is clear that $600+\frac{600}{3}<2600$), 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 answer 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 (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 4

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

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
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

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