AMC 10 · 2011 · #5
Grade 6 rate-ratioPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The three groups are not the same size, so you cannot just average 12, 15, and 10. Naming the number of fifth graders with one variable lets you write every group's size in terms of it. The variable then cancels, so you can even pretend there is just one fifth grader to keep the arithmetic light.
Name the group sizes
Let f be the number of fifth graders. Then there are 2f fourth graders and 4f third graders.
One unknown for the smallest group lets the ratio chain build every other group size from it.
6.RP.A.3Introduce A VariableWrite the overall average
Total minutes is each group's size times its average: 4f × 12 + 2f × 15 + f × 10, over 4f + 2f + f students.
Averaging a mix means pooling all the minutes and all the runners, not averaging the three group averages.
Averaging a mix means pooling all the minutes and all the runners, not averaging the three averages.
▸ Why?
An average is a total shared over a count, so both totals must be rebuilt first.
▸ Why?
A larger group carries more of the total, so each average counts as heavily as its group is big.
Add up the totals
Adding: 48f + 30f + 10f = 88f minutes for 4f + 2f + f = 7f students.
Weighting each average by its group's size is just size times rate for each group, then summed.
5.NBT.B.5Analyze The UnitsDivide and let f cancel
The f cancels in 88f/7f, so the head count never mattered: the average is 88/7, choice (C).
Because both totals scale with the same f, the ratio is fixed, so the actual head count is irrelevant.
5.NF.B.3Solve An Easier Related ProblemTo average groups of different sizes, add up everyone's minutes and divide by everyone, not the three averages.
- Name the group sizes
- Write the overall average
- Add up the totals
- Divide and let f cancel