AMC 8 · 2004 · #20

Easy mode Grade 5

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Problem

Picture a room with some people in it and some chairs. Some people are sitting; the rest are standing.

Here is what we know:

$\frac{2}{3}$ of the people in the room are sitting.
The people who are sitting fill $\frac{3}{4}$ of the chairs.
There are $6$ empty chairs.

How many people are in the room?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: In a room, $\tfrac{2}{3}$ of the people are seated and they fill $\tfrac{3}{4}$ of the chairs. The other $\tfrac{1}{4}$ of the chairs — that is, $6$ chairs — are empty. How many people are in the room?

Givens: $\tfrac{2}{3}$ of the people in the room are seated; The seated people occupy $\tfrac{3}{4}$ of the chairs; There are $6$ empty chairs; Answer choices: (A) $12$, (B) $18$, (C) $24$, (D) $27$, (E) $36$

Unknowns: The total number of people in the room

Understand

Plan

Primary tool: #11 Work Backwards

Secondary: #7 Identify Subproblems

The question asks for the number of people, but the only concrete number we are given is the $6$ empty chairs, which is at the end of the chain. Tool #11 (Work Backwards) says: start from that known piece and undo the fractions one at a time. Tool #7 (Identify Subproblems) breaks the trip in two: first turn "$6$ empty chairs $=$ $\tfrac{1}{4}$ of all chairs" into the total number of chairs, then turn "seated people $=$ $\tfrac{2}{3}$ of all people" into the total number of people. Each subproblem is a one-step Grade 4-5 fraction question.

Execute — Answer: D

#11 Work Backwards 4.NF.B.4 Step 1

Subproblem 1 — find the total number of chairs.
Since $\tfrac{3}{4}$ of the chairs are taken, the remaining $\tfrac{1}{4}$ are empty.
So $6$ empty chairs $=$ $\tfrac{1}{4}$ of the total.
To undo "$\tfrac{1}{4}$ of," multiply by $4$.

$$\tfrac{1}{4} \cdot \text{(total chairs)} = 6 \;\Rightarrow\; \text{total chairs} = 6 \times 4 = 24$$

💡 Grade 4 "multiplying a fraction by a whole number" in reverse: if one part of four equals $6$, four parts equal $4 \times 6 = 24$.

#7 Identify Subproblems 5.NF.B.6 Step 2

Count the seated people.
The chairs that are taken make up $\tfrac{3}{4}$ of the $24$ chairs, and each taken chair holds exactly one seated person.

$$\text{seated people} = \tfrac{3}{4} \times 24 = 18$$

💡 Grade 5 "fraction of a quantity" with whole-number outcome — three of the four equal groups of $6$ chairs are taken, so $3 \times 6 = 18$.

#11 Work Backwards 5.NF.B.7 Step 3

Subproblem 2 — find the total number of people.
The $18$ seated people are $\tfrac{2}{3}$ of everyone in the room, so $\tfrac{1}{3}$ of the people equals half of $18$, which is $9$.
Multiply by $3$ to undo "$\tfrac{1}{3}$ of."

$$\tfrac{2}{3} \cdot \text{(total people)} = 18 \;\Rightarrow\; \tfrac{1}{3} \cdot \text{(total people)} = 9 \;\Rightarrow\; \text{total people} = 3 \times 9 = 27$$

💡 Grade 5 "divide a whole number by a unit fraction": $18 \div \tfrac{2}{3} = 27$, or equivalently split $18$ into two equal parts to get $\tfrac{1}{3}$, then take three of those parts.

#7 Identify Subproblems 5.NF.B.6 Step 4

Read off the answer.

$$27 \;\Rightarrow\; \textbf{(D)}$$

💡 Both subproblems closed cleanly with whole-number answers — a good sign the working-backwards chain was set up correctly.

[1] #11 4.NF.B.4 Subproblem 1 — find the total number of chairs. Since $\tfrac{3}{4}$ of the chai

[2] #7 5.NF.B.6 Count the seated people. The chairs that are taken make up $\tfrac{3}{4}$ of the

[3] #11 5.NF.B.7 Subproblem 2 — find the total number of people. The $18$ seated people are $\tfr

[4] #7 5.NF.B.6 Read off the answer.

Review

Reasonableness: Check by going forward. With $27$ people, the seated group is $\tfrac{2}{3} \times 27 = 18$ and the standing group is $27 - 18 = 9$. With $24$ chairs, the taken chairs are $\tfrac{3}{4} \times 24 = 18$ (matches the $18$ seated people) and the empty chairs are $24 - 18 = 6$ (matches the problem). Eliminations: $12$ and $18$ are too small because $\tfrac{2}{3}$ of $27$ already equals $18$. $24$ is the chair count, not the people count — a classic trap answer. $36$ would force $\tfrac{2}{3} \times 36 = 24$ seated people, more than the $18$ taken chairs allow. Only $27$ is consistent end-to-end.

Alternative: Tool #6 (Guess and Check): the people count must be divisible by $3$ so that $\tfrac{2}{3}$ of it is a whole number, which leaves $(A) 12$, $(D) 27$, $(E) 36$. For each, compute seated $=$ $\tfrac{2}{3} \times \text{people}$ and the chairs needed $=$ $\tfrac{4}{3} \times \text{seated}$, then check empty $=$ $\tfrac{1}{4} \times \text{chairs}$. Only people $= 27$ gives $18$ seated, $24$ chairs, and $6$ empty — answer $(D)$.

CCSS standards used (min grade 5)

4.NF.B.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number (Recognizing that the $6$ empty chairs are $\tfrac{1}{4}$ of the chairs and reversing that to get $4 \times 6 = 24$ total chairs.)
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers (Computing $\tfrac{3}{4} \times 24 = 18$ seated people from the chair count, framed as a fraction-of-a-quantity word problem.)
5.NF.B.7 Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions (Undoing "$\tfrac{2}{3}$ of the people $= 18$" by finding $\tfrac{1}{3} = 9$ and then $3 \times 9 = 27$, which is $18 \div \tfrac{2}{3}$.)

⭐ When the only number you know is at the end of a fraction chain, work backwards: turn $6$ empty chairs into $24$ total chairs, $18$ seated people, and finally $27$ people in the room.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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