AMC 10 · 2005 · #6

Grade 6 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
mean-median-mode-rangeweighted-average identify-subproblemsconvert-to-algebra ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
One group of twenty numbers has an average of thirty. A separate group of thirty numbers has an average of twenty. Put all fifty numbers together and find the average of the whole collection.

Pick an answer.

(A)
23
(B)
24
(C)
25
(D)
26
(E)
27

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Averages cannot be added together, but totals can, so Tool #16 (Change Focus) shifts attention from the two averages to the two hidden totals. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: the total of the first group, the total of the second group, and the average of everything combined. Tool #3 (Eliminate Possibilities) guards the main trap, because simply averaging the two averages 30 and 20 gives 25, which is choice (C) but is wrong since the two groups are not the same size.

1STEP 1

Turn the first average into a total

An average is a total shared out evenly, so total = average × count. The first group: 20 numbers averaging 30, so its total is 600.

sum₁ = 20 × 30 = 600
2STEP 2

Turn the second average into a total

Same move on the second group: 30 numbers averaging 20 give a total of 600 — equal to the first, despite the different sizes.

sum₂ = 30 × 20 = 600
3STEP 3

Add the totals and count all the numbers

Real sums add directly: grand total 1200 over a count of 50 — exactly what a combined average needs.

grand total = 600 + 600 = 1200, count = 20 + 30 = 50
4STEP 4

Divide to get the combined average

Divide: 1200 ÷ 50 = 24. Averaging the two averages gives 25, choice (C) — wrong, since the bigger group averages 20 and pulls down.

1200/50 = 24 → (B)
Answer
24
The combined average must land somewhere between the two group averages 20 and 30, and 24 does. Because the larger group (the 30 numbers) averages the smaller value 20, the result should sit below the plain midpoint 25, and 24 is indeed just under 25. If the group sizes were swapped, the pull would go the other way and the average would rise above 25 instead. Choice (C) 25 is exactly the midpoint you would get by wrongly averaging the two averages, which is why it is the tempting wrong answer.
💡Key takeaway

You cannot average two averages when the groups are different sizes; turn each average back into a total, add the totals, and divide by how many numbers there are in all.

  • Turn the first average into a total
  • Turn the second average into a total
  • Add the totals and count all the numbers
  • Divide to get the combined average

More like this

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