AMC 10 · 2011 · #4

Grade 6 rate-ratio
weighted-averageratio-proportionmean-median-mode-range convert-to-algebra ↑ Prerequisites: ratio-proportion
📏 Medium solution 💡 2 insights
Problem
Three groups have known averages and group sizes that double down the list. Find the overall average.

Pick an answer.

(A)
12
(B)
$\frac{37}{3}$
(C)
$\frac{88}{7}$
(D)
13
(E)
14
How to solve
Strategy Introduce a Variable

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.

1STEP 1

Name the group sizes

One letter names all three group sizes.

fifth = f, fourth = 2f, third = 4f
2STEP 2

Write the overall average

The average is a total over a count.

average = (4f × 12 + 2f × 15 + f × 10)/(4f + 2f + f)
3STEP 3

Add up the totals

Both sums carry the same unknown.

48f + 30f + 10f = 88f, 4f + 2f + f = 7f
4STEP 4

Divide and let f cancel

It cancels, leaving 88/7, choice (B).

88f/7f = 88/7
Answer
88/7
The overall average 88/7 ≈ 12.57 sits between the smallest group average (10) and the largest (15), which any true average must. It leans toward 12 because the biggest group (third graders, 4f of them) runs 12 minutes, pulling the mean close to 12. That matches choice (C) and rules out 12 exactly (A) or 13+ (D, E).
💡Key takeaway

To 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