AMC 10 · 2021 · #2

Grade 4 arithmetic
ratio-proportionlinear-equations-one-varmulti-digit-arithmetic physical-representationidentify-subproblemsguess-and-check ↑ Prerequisites: ratio-proportion
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Problem
Portia's school has 3 times the students of Lara's school. Together they have 2600 students. Find how many students attend Portia's school.

Pick an answer.

(A)
~600
(B)
~650
(C)
~1950
(D)
~2000
(E)
~2050

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

"3 times as many" begs for a tape diagram (tool #1): draw one box for Lara and three identical boxes for Portia. That makes the four boxes together equal to 2600, so one box is 2600 ÷ 4 = 650 and Portia gets 3 boxes. Tool #6 (Guess and Check) and tool #3 (Eliminate) confirm: the answer must be more than half of 2600, knocking out (A), (B); and the matching small box value 650 shows up as choice (B) for Lara — confirming Portia's 1950.

1STEP 1

Draw each school as identical boxes: Lara is one box, Portia is three — so together they make four equal boxes.

Lara: □ Portia: □ □ □ Total: □ □ □ □ = 2600
2STEP 2

Four equal boxes total 2600, so each box is 2600 ÷ 4 = 650 students — that is Lara's school.

2600 ÷ 4 = 650
3STEP 3

Portia's school is 3 boxes: 3 × 650 = 1950 students.

3 × 650 = 1950
4STEP 4

Check: 1950 is choice (C), and it beats half of 2600, matching that Portia's share is the larger one.

1950 + 650 = 2600 ✓, 1950 = 3 × 650 ✓ → (C)
Answer
~1950
Portia's share is the larger one, so the answer must beat half of 2600 = 1300. Choices (A) 600 and (B) 650 are too small. (C) 1950 sits at three-quarters of 2600, which is exactly what "3 of 4 equal boxes" predicts. The other big choices (D) 2000 and (E) 2050 do not make 4 equal boxes with the rest.
💡Key takeaway

This AMC 10 problem only needs Grade 4 "3 times as many" tape diagrams you already know — split 2600 into 4 equal boxes, give Portia 3 of them, and the answer is 1950.