Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 10A Problems/Problem 2"
(→‎Video Solution 1 (Very Fast & Simple))
 
(33 intermediate revisions by 11 users not shown)
Line 4: Line 4:
 
<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 2 (Setting Variables)===
  
−
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://youtu.be/qNf6SiIpIsk?t=119
  
−
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.
 
−
 
−
~MRENTHUSIASM
 
−
 
−
==Video Solution #1(Setting Variables) ==
 
−
 
−
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 by OmegaLearn===
 
https://youtu.be/xXx0iP1tn8k
 
https://youtu.be/xXx0iP1tn8k
  
−
- pi_is_3.14
+
~pi_is_3.14
 +
 
 +
===Video Solution 5===
 +
https://youtu.be/GwwDQYqptlQ
  
−
==Video Solution (Simple)==
+
~savannahsolver
−
https://youtu.be/DOtysU-a1B4
+
 
 +
===Video Solution 6===
 +
https://youtu.be/50CThrk3RcM?t=66
 +
 
 +
~IceMatrix
 +
 
 +
===Video Solution 7 (Problems 1-3)===
 +
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 09:06, 11 July 2024

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 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

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)
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_2&oldid=222058"