AMC 10 · 2013 · #5

Grade 6 rate-ratio
weighted-averagemean-median-mode-range identify-subproblemsdimensional-analysis ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 1 insight
Problem
Two groups of different sizes have known average ages. Find the average age of everyone.

Pick an answer.

(A)
22
(B)
23.25
(C)
24.75
(D)
26.25
(E)
28
How to solve
Strategy Identify Subproblems

To average two groups together I first need each group's total age. I break the work into three subproblems: the total age of the kids, the total age of the parents, then one division of the combined total by the combined head count. Tracking the units (years of age per person) keeps that final division honest.

1STEP 1

Averages hide totals

Every average hides a total.

average = total/count
2STEP 2

Total age of the fifth-graders

The first group's total is 363.

33 × 11 = 363
3STEP 3

Total age of the parents

The second group's total is 1815.

55 × 33 = 1815
4STEP 4

Combine totals and people

Both totals and both counts combine.

363 + 1815 = 2178, 33 + 55 = 88
5STEP 5

Divide once to get the average

One division gives 24.75, choice (B).

2178/88 = 24.75
Answer
24.75
The result 24.75 lands between 11 and 33, as any blend of the two averages must. Since there are more parents (55) than fifth-graders (33), the blend should lean toward the parents' 33 rather than sit at the plain midpoint of 22 — and 24.75 is indeed above 22. Choice (A) 22 is exactly the trap you fall into by wrongly averaging the two averages.
💡Key takeaway

You cannot average two averages when the groups are different sizes — turn each average back into a total, add everything up, then divide just once.

  • Averages hide totals
  • Total age of the fifth-graders
  • Total age of the parents
  • Combine totals and people
  • Divide once to get the average