AMC 10 · 2021 · #2

Easy mode Grade 4
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Problem

Picture two high schools. One belongs to Portia, and the other belongs to Lara.

Portia's school has 33 times as many students as Lara's school. Put together, the two schools have 26002600 students in total.

How many students go to Portia's school?

Pick an answer.

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

⭐ 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 $1950$.