AMC 8 · 2007 · #7
Grade 6 arithmeticProblem
The average age of people in a room is years. An -year-old person leaves
the room. What is the average age of the four remaining people?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Five people in a room have an average age of $30$. One of them, who is $18$, walks out. What is the average age of the four people who stay?
Givens: There are $5$ people in the room; Their average age is $30$ years; An $18$-year-old leaves; Answer choices: (A) $25$, (B) $26$, (C) $29$, (D) $33$, (E) $36$
Unknowns: The average age of the $4$ remaining people
Understand
Restated: Five people in a room have an average age of $30$. One of them, who is $18$, walks out. What is the average age of the four people who stay?
Givens: There are $5$ people in the room; Their average age is $30$ years; An $18$-year-old leaves; Answer choices: (A) $25$, (B) $26$, (C) $29$, (D) $33$, (E) $36$
Plan
Primary tool: #7 Identify Subproblems
Tool #7 (Identify Subproblems) fits because averages do not subtract directly — you cannot just take $30 - 18$. Instead the question splits into three small subproblems, each one a single Grade 6 step: (a) recover the original sum of ages from the original average, (b) subtract the leaver's age to get the new sum, (c) divide the new sum by the new count. No variable or equation is needed — the mean formula run forward and backward is enough.
Execute — Answer: D
6.SP.B.5 Step 1 - Recover the original total.
- The mean formula says sum $=$ average $\times$ count.
- With $5$ people averaging $30$, the total of their ages is $5 \times 30$.
💡 Grade 6 mean formula in reverse: knowing the average and the count pins down the total.
4.OA.A.3 Step 2 - Subtract the leaver.
- The $18$-year-old walks out, so the total drops by exactly $18$.
- Nobody else's age changes.
💡 Only the total changes here — the count change is handled in the next step.
6.SP.B.5 Step 3 - Re-average.
- Divide the new total by the new count of $4$ people.
💡 Same mean formula, applied forward this time: sum $\div$ count.
6.SP.B.5 Recover the original total. The mean formula says sum $=$ average $\times$ count 4.OA.A.3 Subtract the leaver. The $18$-year-old walks out, so the total drops by exactly 6.SP.B.5 Re-average. Divide the new total by the new count of $4$ people. Review
Reasonableness: Check the direction first. The person who left was $18$, well below the group average of $30$, so pulling them out should raise the average — and $33 > 30$, which matches. Verify by running it the other way: if the $4$ remaining people average $33$, their total is $4 \times 33 = 132$. Add back the $18$-year-old to get $132 + 18 = 150$, and $150 / 5 = 30$, the original average. The numbers close the loop.
Alternative: Tool #3 (Write an Equation) gives the same answer in one line. Let $N$ be the new average. The total of all $5$ ages can be written two ways: as $5 \times 30 = 150$ from the original average, and as (sum of the $4$ who stayed) $+ 18 = 4N + 18$ from the new average plus the leaver. Setting them equal, $4N + 18 = 150$, so $4N = 132$ and $N = 33$.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the mean formula both ways: average $\times$ count $=$ sum to recover the original total of $150$, and sum $\div$ count to compute the new average $132/4 = 33$.)4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Subtracting the leaver's age from the original total, $150 - 18 = 132$, as the middle step of the multi-step plan.)
⭐ Averages do not subtract. To handle a "someone leaves" problem, go back to the total, take the leaver's age out, then divide by the new count.
⭐ Averages do not subtract. To handle a "someone leaves" problem, go back to the total, take the leaver's age out, then divide by the new count.