2021 AMC 10A Problem 2

"Three times as many" is easiest to handle by thinking in equal shares.

1. Call the size of Lara's school one share. Portia's school is three of those shares.
2. Together the two schools are $1 + 3 = 4$ shares, and that is $2600$ students.
3. One share is $2600 \div 4 = 650$ students. This is Lara's school.
4. Portia's school is three shares: $3 \times 650 = 1950$ students.
5. Check: $1950 + 650 = 2600$, and $1950$ is three times $650$.

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

When one amount is a number of times another, the total is made of equal shares: here $1$ share plus $3$ shares. Divide the total by the number of shares to find one share, then multiply. Using a letter works the same way: if Lara's school has $L$ students, then $L + 3L = 4L = 2600$.

Test the choices. For each one, subtract it from $2600$ to get Lara's school, and see whether Portia's number is three times Lara's. For $1950$, Lara's school has $2600 - 1950 = 650$, and $3 \times 650 = 1950$. It fits. For $2000$, Lara's school would have $600$, and $3 \times 600 = 1800$, not $2000$.