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

Revision as of 07:48, 12 February 2021

Problem 2

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

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. $p=3q$ $p+q=2600$ Substituting $p$ with $3q$ we get $4q=2600$ . Solving for $q$ , we get $q=650$ . 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 (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 $\boxed{\textbf{(C)} ~1950}.$

Video Solution

https://youtu.be/xXx0iP1tn8k

- pi_is_3.14

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.

@@ Line 16: / Line 16: @@
 ~MRENTHUSIASM
+==Solution 3 (Quick Inspection)==
+The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)}</math>, <math>\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>
 ==Video Solution==

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