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2011 AMC 10A Problem 5

2011 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 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 grades are in the ratio 4:2:1, so take 4, 2 and 1 students and compute a weighted average: (48 + 30 + 10)/7.

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

Why this works

An overall average is total divided by count, so groups of different sizes must be weighted by size. When only ratios are given, scaling all group sizes by the same factor scales the total and the count equally, so choosing $4, 2, 1$ loses nothing. Choice (B) is exactly the unweighted average, placed there as bait.

The trap

Averaging the three grade averages equally, (12 + 15 + 10)/3 = 37/3, ignoring that the grades have different sizes.

Common mistakes

Averaging the three grade averages equally, (12 + 15 + 10)/3 = 37/3, ignoring that the grades have different sizes.
Reversing the ratio (giving the fifth grade the most students), which produces $\frac{94}{7}$ , not among the choices, and then guessing.

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