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

Revision as of 21:42, 11 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

Video Solution

https://youtu.be/xXx0iP1tn8k

- pi_is_3.14

@@ Line 13: / Line 13: @@
 ==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{\text{(C) }1950}.</cmath>
+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>
 ~MRENTHUSIASM

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