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2011 AMC 10A Problem 5

2011 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: October 28, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 2 min · position P1-10

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?

A) $12$
B) $\frac{37}{3}$
C) $\frac{88}{7}$
D) $13$
E) $14$

Answer

Key insight

The grade sizes are in the ratio 4 : 2 : 1, so pretend there are 4, 2 and 1 students. Add up all their minutes and divide by 7.

Solution

The problem gives no actual numbers of students, only how the grades compare. So choose small numbers that fit.

Start with the smallest grade. Say there is $1$ fifth grader.
There are twice as many fourth graders as fifth graders, so $2$ fourth graders.
There are twice as many third graders as fourth graders, so $4$ third graders.
That makes $4 + 2 + 1 = 7$ students.
Now add up the minutes run per day by everyone:

Grade	Students	Minutes each	Total minutes
3rd	$4$	$12$	$48$
4th	$2$	$15$	$30$
5th	$1$	$10$	$10$
All	$7$		$88$

The average is total minutes divided by number of students: $\dfrac{88}{7}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{88}{7}}$ .

Why this works

An overall average is always "total divided by how many". A big grade adds more to the total than a small one, so you cannot just average the three grade averages. When only the ratio of the group sizes is given, any numbers in that ratio give the same answer (doubling every group doubles both the total and the count), so pick the smallest ones.

The trap

Averaging the three grade averages as if the grades were the same size: (12 + 15 + 10) ÷ 3 = 37/3, which is choice (B).

Concise solution

Only the ratio of the grade sizes matters, so pick the smallest convenient numbers: $1$ fifth grader, $2$ fourth graders, and $4$ third graders, $7$ students in all.

Total minutes run per day:
$4 \cdot 12 + 2 \cdot 15 + 1 \cdot 10 = 48 + 30 + 10 = 88.$

The average over all $7$ students is $\dfrac{88}{7}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{88}{7}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Prealgebra, Chapter 13: Data and Statistics
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer