AMC 10 · 2021 · #2

Grade 4 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up

ratio-proportionlinear-equations-one-varmulti-digit-arithmetic physical-representationidentify-subproblemsguess-and-check ↑ Prerequisites: ratio-proportion

📏 Short solution 💡 1 insight

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Problem

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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Portia's students $= 3 \times$ Lara's students; Portia's students $+$ Lara's students $= 2600$; Answer choices: (A) $600$, (B) $650$, (C) $1950$, (D) $2000$, (E) $2050$

Unknowns: The number of students at Portia's school

Understand

Restated: 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.

Givens: Portia's students $= 3 \times$ Lara's students; Portia's students $+$ Lara's students $= 2600$; Answer choices: (A) $600$, (B) $650$, (C) $1950$, (D) $2000$, (E) $2050$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #6 Guess and Check, #3 Eliminate Possibilities

"$3$ 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 $2600 \div 4 = 650$ 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 $1950$.

Execute — Answer: C

#1 Draw a Diagram 4.OA.A.2 Step 1

Picture the schools as identical boxes.
Lara's school is one box.
Portia's school is three boxes the same size.
Together that is four equal boxes.

Lara: $\square$ \quad Portia: $\square\,\square\,\square$ \quad Total: $\square\,\square\,\square\,\square = 2600$

💡 "$3$ times as many" is a Grade 4 multiplicative comparison — the bar model turns the comparison into countable boxes.

#1 Draw a Diagram 4.NBT.B.6 Step 2

Four equal boxes total $2600$, so each box is $2600 \div 4 = 650$ students.
That is the size of Lara's school.

$$2600 \div 4 = 650$$

💡 Splitting a four-digit total evenly among $4$ groups is Grade 4 long division.

#1 Draw a Diagram 4.NBT.B.5 Step 3

Portia's school is $3$ boxes, so $3 \times 650 = 1950$ students.

$$3 \times 650 = 1950$$

💡 Multiplying a three-digit number by a one-digit number is Grade 4 "multi-digit by one-digit".

#3 Eliminate Possibilities 4.OA.A.3 Step 4

Check against the choices.
$1950$ is choice (C) — and it sits comfortably above half of $2600$, matching the constraint that Portia's share is the larger one.

$$1950 + 650 = 2600 \;\checkmark, \;\; 1950 = 3 \times 650 \;\checkmark \;\Rightarrow\; \textbf{(C)}$$

💡 Plugging the candidate back into the two given relationships and seeing them both hold is Grade 4 "multi-step word problem with reasonableness check".

[1] #1 4.OA.A.2 Picture the schools as identical boxes. Lara's school is one box. Portia's schoo

[2] #1 4.NBT.B.6 Four equal boxes total $2600$, so each box is $2600 \div 4 = 650$ students. That

[3] #1 4.NBT.B.5 Portia's school is $3$ boxes, so $3 \times 650 = 1950$ students.

[4] #3 4.OA.A.3 Check against the choices. $1950$ is choice (C) — and it sits comfortably above

Review

Reasonableness: 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.

Alternative: Tool #6 (Guess and Check) directly on the choices: test choice (C). If Portia $= 1950$, then Lara $= 2600 - 1950 = 650$, and $1950 \div 650 = 3$ exactly. Both conditions satisfied — pick (C).

CCSS standards used (min grade 4)

4.OA.A.2 Multiply or divide to solve word problems involving multiplicative comparison (Modeling "$3$ times as many" as one box for Lara and three identical boxes for Portia.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Computing $3 \times 650 = 1950$ to get Portia's enrollment.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing the total $2600 \div 4 = 650$ to find one box's worth of students.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Verifying $1950 + 650 = 2600$ and $1950 = 3 \times 650$ in one combined check.)

⭐ 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 $1950$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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