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2021 AMC 10A Problem 5

2021 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

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

The quiz scores of a class with $k > 12$ students have a mean of $8$ . The mean of a collection of $12$ of these quiz scores is $14$ . What is the mean of the remaining quiz scores in terms of $k$ ?

A)
$\frac{14-8}{k-12}$
B)
$\frac{8k-168}{k-12}$
C)
$\frac{14}{12} - \frac{8}{k}$
D)
$\frac{14(k-12)}{k^2}$
E)
$\frac{14(k-12)}{8k}$

Answer

Key insight

Work with totals, not means: the class sums to 8k, the twelve scores sum to 168, and the remaining k - 12 scores share what is left.

Solution

Convert every mean into a total. The whole class of $k$ students has total score $8k$ . The $12$ selected students have total $12 \cdot 14 = 168$ .

The remaining $k - 12$ students therefore have total $8k - 168$ , and their mean is
$\frac{8k - 168}{k - 12}.$

The answer is $\boxed{\textbf{(B)}\ \frac{8k-168}{k-12}}$ .

Why this works

Means cannot be added or subtracted directly, but totals can. The reliable routine for any "mean of the rest" question is: total of all, minus total of the part, divided by the count of the rest.

Alternative approach

Plug in a value: with $k = 13$ , the thirteenth score is $8 \cdot 13 - 168 = -64$ , so the answer should be $-64$ . Only choice (B) gives $\frac{104 - 168}{1} = -64$ .

The trap

Averaging the two means (subtracting 14 from 8 or similar) instead of subtracting totals, which produces choice (A).

Common mistakes

Averaging the two means (subtracting 14 from 8 or similar) instead of subtracting totals, which produces choice (A).
Dividing the leftover total by $k$ instead of by $k - 12$ .

Techniques

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