AMC 8 · 2008 · #10
Grade 6 arithmeticProblem
The average age of the 6 people in Room A is 40. The average age of the 4 people in Room B is 25. If the two groups are combined, what is the average age of all the people?
Pick an answer.
AMC 8 2008 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: Room A has $6$ people whose ages average $40$. Room B has $4$ people whose ages average $25$. If everyone gathers in one room, what is the average age of the combined group of $10$ people?
Givens: Room A: $6$ people, average age $40$; Room B: $4$ people, average age $25$; Answer choices: (A) $32.5$, (B) $33$, (C) $33.5$, (D) $34$, (E) $35$
Unknowns: The average age of all $10$ people when the two rooms are combined
Understand
Restated: Room A has $6$ people whose ages average $40$. Room B has $4$ people whose ages average $25$. If everyone gathers in one room, what is the average age of the combined group of $10$ people?
Givens: Room A: $6$ people, average age $40$; Room B: $4$ people, average age $25$; Answer choices: (A) $32.5$, (B) $33$, (C) $33.5$, (D) $34$, (E) $35$
Plan
Primary tool: #16 Transform the Structure
Secondary: #7 Identify Subproblems
You cannot just average $40$ and $25$ — the two rooms have different sizes, so a plain $(40+25)/2$ throws away that fact. Tool #16 (Transform the Structure) says: turn each average back into a total. Once both rooms are described by their total age (sum), the combined room is just sum-of-sums divided by count-of-counts. Tool #7 (Identify Subproblems) breaks the work into two clean steps — find each room's total, then merge.
Execute — Answer: D
6.SP.B.5 Step 1 - Subproblem 1: find the total age in Room A.
- The average times the count gives the total, so multiply $40$ by $6$.
💡 If the average age is $40$, you can pretend every person in Room A is $40$ years old. Six pretend-$40$s add to $240$.
6.SP.B.5 Step 2 Subproblem 2: find the total age in Room B the same way — average times count.
💡 Four pretend-$25$s add to $100$.
5.NBT.B.5 Step 3 - Combine the rooms.
- The combined total age is the sum of the two room totals, and the combined count is $6+4=10$.
💡 Pour both rooms into one and just count everyone and add all the ages.
6.SP.B.5 Step 4 Apply the average formula one more time to the combined group.
💡 Sum $\div$ count gives the new average. Dividing by $10$ just shifts the decimal one place.
6.SP.B.5 Subproblem 1: find the total age in Room A. The average times the count gives th 6.SP.B.5 Subproblem 2: find the total age in Room B the same way — average times count. 5.NBT.B.5 Combine the rooms. The combined total age is the sum of the two room totals, and 6.SP.B.5 Apply the average formula one more time to the combined group. Review
Reasonableness: The answer $34$ sits between $25$ (Room B's average) and $40$ (Room A's average), which it has to — a combined average can never escape the range of the two parts. It also sits closer to $40$ than to $25$, which fits because Room A has more people ($6$ vs $4$), so it pulls the average toward $40$. A quick gut check: midpoint of $25$ and $40$ is $32.5$; we should land above $32.5$ because the bigger group is the older one, and indeed $34 > 32.5$.
Alternative: Tool #4 (Introduce a Variable) for a weighted-average formula. Let $\bar{x}$ be the combined average. Then $\bar{x} = \dfrac{6 \cdot 40 + 4 \cdot 25}{6 + 4} = \dfrac{340}{10} = 34$. This is the same calculation in one line — the weighted-average shortcut for two groups.
CCSS standards used (min grade 6)
5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Computing each room's total age ($6 \times 40 = 240$ and $4 \times 25 = 100$) and the combined total.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the mean formula (sum $\div$ count) backwards to recover each room's total, and then forwards to compute the combined average $340 \div 10 = 34$.)
⭐ When two groups of different sizes are combined, you can't just average the averages. Turn each average back into a total, add the totals, and divide by the new headcount — the bigger group always pulls harder.
⭐ When two groups of different sizes are combined, you can't just average the averages. Turn each average back into a total, add the totals, and divide by the new headcount — the bigger group always pulls harder.
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