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2021 AMC 10A Problem 2

2021 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$600$
B)
$650$
C)
$1950$
D)
$2000$
E)
$2050$

Answer

Key insight

The two schools split 2600 students in the ratio 3:1, so Portia's school has three of the four equal parts.

Solution

Let Lara's school have $L$ students. Then Portia's school has $3L$ students, and the total is
$L + 3L = 4L = 2600,$
so $L = 650$ . Portia's school therefore has $3 \cdot 650 = 1950$ students.

The answer is $\boxed{\textbf{(C)}\ 1950}$ .

Why this works

"Three times as many" makes the total four equal parts, so one part is $2600/4$ . Writing the smaller quantity as the variable keeps the algebra to a single equation with no fractions.

Alternative approach

Check the choices: only $1950$ leaves a remainder ( $650$ ) that is exactly one third of it.

The trap

Reporting Lara's share, 650, or dividing 2600 by 3 instead of by 4.

Common mistakes

Reporting Lara's share, 650, or dividing 2600 by 3 instead of by 4.
Setting Portia's count as the variable $P$ and writing $P + 3P = 2600$ , which reverses the roles and gives $650$ .

Techniques

Set up the equation/formula and compute; no special trick needed