Competition · AMC preparation · step 4 of 4
AMC 8 · 2008 · #10
Grade 6 arithmeticPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Total the ages in Room A
Room A's total age is its average times its count: 40 across 6 people gives 240.
If the average age is 40, you can pretend every person in Room A is 40 years old. Six pretend-40s add to 240.
The 6 people in Room A, whose ages average 40, have a combined total age of 6 × 40 = 240.
▸ Why?
Saying the average age is 40 just means the total age has been shared equally among the 6 people, so the total is recovered by multiplying that shared amount back by the count: 6 × 40 = 240.
Total the ages in Room B
Room B's total age the same way: 25 across 4 people gives 100.
Four pretend-25s add to 100.
6.SP.B.5Identify SubproblemsCombine both rooms
Pour both rooms together: the totals add to 340 across 10 people.
Pour both rooms into one and just count everyone and add all the ages.
5.NBT.B.5Change Focus Count The ComplementAverage the combined group
Apply the average formula once more to the whole group: 340 over 10 is 34.
Sum ÷ count gives the new average. Dividing by 10 just shifts the decimal one place.
6.SP.B.5Change Focus Count The ComplementWhen 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.
- Total the ages in Room A
- Total the ages in Room B
- Combine both rooms
- Average the combined group
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