AMC 10 · 2013 · #6
Grade 6 arithmeticThe average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?
Pick an answer.
AMC 10 2013 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: There are 33 fifth-graders whose ages average 11, and 55 parents whose ages average 33. Find the single average age of everyone in the combined group of 88 people.
Givens: 33 fifth-graders with an average age of 11; 55 parents with an average age of 33
Unknowns: The average age of all 88 people taken together
Understand
Restated: There are 33 fifth-graders whose ages average 11, and 55 parents whose ages average 33. Find the single average age of everyone in the combined group of 88 people.
Givens: 33 fifth-graders with an average age of 11; 55 parents with an average age of 33
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
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.
Execute — Answer: C
6.SP.A.3 Step 1 Averages hide totals
- An average by itself cannot be merged with another average unless the two groups are the same size.
- Each average is really a total shared out evenly: total = average x group size.
- Because these groups differ in size (33 versus 55), I must rebuild each total before combining.
💡 An average is a total split evenly, so to merge groups you first need the totals back.
5.NBT.B.5 Step 2 Total age of the fifth-graders
The 33 fifth-graders each average 11 years, so their ages add up to 33 times 11.
💡 Multiplying the average by the head count rebuilds the group's whole pile of years.
5.NBT.B.5 Step 3 Total age of the parents
The 55 parents each average 33 years, so their ages add up to 55 times 33.
💡 Same move as the kids: average times count gives that group's total years.
4.NBT.B.4 Step 4 Combine totals and people
Pour both age totals into one and count all the heads together: the ages sum to 363 + 1815 = 2178 years, spread across 33 + 55 = 88 people.
💡 One combined group has one combined total and one combined count.
6.NS.B.3 Step 5 Divide once to get the average
- Share the combined 2178 years over all 88 people.
- Dividing gives 24.75 years per person, so the answer is (C).
💡 Total age shared over every person is age per person, which is exactly the average.
6.SP.A.3 An average by itself cannot be merged with another average unless the two groups 5.NBT.B.5 The 33 fifth-graders each average 11 years, so their ages add up to 33 times 11. 5.NBT.B.5 The 55 parents each average 33 years, so their ages add up to 55 times 33. 4.NBT.B.4 Pour both age totals into one and count all the heads together: the ages sum to 6.NS.B.3 Share the combined 2178 years over all 88 people. Dividing gives 24.75 years per Review
Reasonableness: 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.
Alternative: Set it up as one weighted average from the start: (33 x 11 + 55 x 33) / (33 + 55). This is the same arithmetic, just written in a single fraction instead of separate subproblems.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Knowing an average equals total divided by count, so two groups combine by their totals, not by averaging their averages)5.NBT.B.5Fluently multiply multi-digit whole numbers (Turning each group's average and size back into its total age (33 x 11 and 55 x 33))4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the two age totals and the two group sizes into one total and one head count)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Dividing the combined total age by the number of people to get 24.75)
⭐ 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.
⭐ 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.
More like this
Same archetype — closest grade level first.