AMC 10 · 2011 · #5

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

weighted-averageratio-proportionmean-median-mode-range convert-to-algebra ↑ Prerequisites: ratio-proportion

📏 Medium solution 💡 2 insights

Problem

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12$ , $15$ , and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

Pick an answer.

(A)

(B)

$\frac{37}{3}$

(C)

$\frac{88}{7}$

(D)

(E)

AMC 10 2011 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.

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Toolkit + CCSS Solution

Understand

Restated: Third, fourth, and fifth graders run averages of 12, 15, and 10 minutes per day. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. Find the overall average minutes run per day across all these students.

Givens: Third graders average 12 minutes/day; Fourth graders average 15 minutes/day; Fifth graders average 10 minutes/day; Third graders = 2 \times fourth graders; Fourth graders = 2 \times fifth graders

Unknowns: The average number of minutes run per day by all the students combined

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #9 Solve an Easier Related Problem, #8 Analyze the Units

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.

Execute — Answer: C

#4 Introduce a Variable 6.RP.A.3 Step 1

Name the group sizes

Let f be the number of fifth graders.
There are twice as many fourth graders as fifth graders, so there are 2f fourth graders.
There are twice as many third graders as fourth graders, so there are 2 \times 2f = 4f third graders.

$$\text{fifth} = f,\quad \text{fourth} = 2f,\quad \text{third} = 4f$$

💡 One unknown for the smallest group lets the ratio chain build every other group size from it.

#8 Analyze the Units 6.SP.B.5 Step 2

Write the overall average

An overall average is the total minutes run by everyone divided by the total number of students.
Each group contributes (its size) \times (its average minutes), so total minutes = 4f \times 12 + 2f \times 15 + f \times 10, and total students = 4f + 2f + f.

$$\text{average} = \dfrac{4f \times 12 + 2f \times 15 + f \times 10}{4f + 2f + f}$$

💡 Averaging a mix means pooling all the minutes and all the runners, not averaging the three group averages.

#8 Analyze the Units 5.NBT.B.5 Step 3

Add up the totals

Multiply each group size by its average and add: 4f \times 12 = 48f, 2f \times 15 = 30f, and f \times 10 = 10f, giving 88f total minutes.
The students add to 4f + 2f + f = 7f.

$$48f + 30f + 10f = 88f,\qquad 4f + 2f + f = 7f$$

💡 Weighting each average by its group's size is just size times rate for each group, then summed.

#9 Solve an Easier Related Problem 5.NF.B.3 Step 4

Divide and let f cancel

The average is \dfrac{88f}{7f}.
The f in the top and bottom cancel, so the answer does not depend on how many students there actually are.
This is the same as pretending there is exactly one fifth grader (f = 1).
The average is \dfrac{88}{7}, which is choice (C).

$$\dfrac{88f}{7f} = \dfrac{88}{7}$$

💡 Because both totals scale with the same f, the ratio is fixed, so the actual head count is irrelevant.

[1] #4 6.RP.A.3 Let f be the number of fifth graders. There are twice as many fourth graders as

[2] #8 6.SP.B.5 An overall average is the total minutes run by everyone divided by the total num

[3] #8 5.NBT.B.5 Multiply each group size by its average and add: 4f \times 12 = 48f, 2f \times 1

[4] #9 5.NF.B.3 The average is \dfrac{88f}{7f}. The f in the top and bottom cancel, so the answe

Review

Reasonableness: The overall average \frac{88}{7} \approx 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).

Alternative: Instead of a variable, use the smallest whole-number counts that fit the ratios: 1 fifth grader, 2 fourth graders, 4 third graders. Then total minutes = 4(12) + 2(15) + 1(10) = 88 over 7 students, giving \frac{88}{7} directly. This is the 'assume one fifth grader' shortcut, and it lands on the same value.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Turning 'twice as many' into the group sizes f, 2f, and 4f)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Setting up the overall (weighted) average as total minutes divided by total students)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing each group's total minutes and summing to 88f over 7f)
5.NF.B.3 Interpret a fraction as division of the numerator by the denominator (Reading 88f divided by 7f as the fraction 88/7 after the f cancels)

⭐ To average groups of different sizes, add up everyone's minutes and divide by everyone, not the three averages.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Name the group sizes

Write the overall average

Add up the totals

Divide and let f cancel

Review

CCSS standards used (min grade 6)

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