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2021 AMC 10B Problem 6

2021 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 2 of 5 · about 2 min · position P1-10

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{ 3}{4}$ . What is the mean of the scores of all the students?

A)
$74$
B)
$75$
C)
$76$
D)
$77$
E)
$78$

Answer

Key insight

Take 3 and 4 students: the total 3·84 + 4·70 = 532 over 7 students gives 76; the answer is a weighted mean.

Solution

Only the ratio of class sizes matters, so suppose the morning class has $3$ students and the afternoon class has $4$ .

Total of the morning scores: $3 \cdot 84 = 252$ . Total of the afternoon scores: $4 \cdot 70 = 280$ . Combined total $252 + 280 = 532$ over $7$ students:
$\frac{532}{7} = 76.$

The answer is $\boxed{\textbf{(C)}\ 76}$ .

Why this works

A combined mean is a weighted average, weighted by class sizes. Scaling both classes by the same factor scales every total by that factor and leaves the mean unchanged, so picking the smallest sizes matching the ratio is always legitimate.

Alternative approach

Deviations from the midpoint: the combined mean sits $\frac{4}{7}$ of the way from $84$ toward $70$ (weights $3 : 4$ ). The gap is $14$ , and $\frac{4}{7} \cdot 14 = 8$ , so the mean is $84 - 8 = 76$ .

The trap

Averaging 84 and 70 to get 77, ignoring that the afternoon class is larger and pulls the mean below the midpoint.

Common mistakes

Averaging 84 and 70 to get 77, ignoring that the afternoon class is larger and pulls the mean below the midpoint.
Weighting backwards (giving the morning class weight $4$ ), which yields $78$ .

Techniques

Set up the equation/formula and compute; no special trick needed