AMC 8 · 2004 · #20
Grade 5 arithmeticProblem
Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are 6 empty chairs, how many people are in the room?
Pick an answer.
AMC 8 2004 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.
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
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) $36$
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
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 $4$.
💡 Grade 4 "multiplying a fraction by a whole number" in reverse: if one part of four equals $6$, four parts equal $4 \times 6 = 24$.
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.
💡 Grade 5 "fraction of a quantity" with whole-number outcome — three of the four equal groups of $6$ chairs are taken, so $3 \times 6 = 18$.
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."
💡 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.
5.NF.B.6 Step 4 Read off the answer.
💡 Both subproblems closed cleanly with whole-number answers — a good sign the working-backwards chain was set up correctly.
4.NF.B.4 Subproblem 1 — find the total number of chairs. Since $\tfrac{3}{4}$ of the chai 5.NF.B.6 Count the seated people. The chairs that are taken make up $\tfrac{3}{4}$ of the 5.NF.B.7 Subproblem 2 — find the total number of people. The $18$ seated people are $\tfr 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.4Apply 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.6Solve 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.7Apply 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.
⭐ 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.
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