2021 AMC 10A Spring Problem 5

Question

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?

\dfrac{14-8}{k-12}

\dfrac{8k-168}{k-12}

\dfrac{14}{12} - \dfrac{8}{k}

\dfrac{14(k-12)}{k^2}

\dfrac{14(k-12)}{8k}

Convert each mean into a total score

Subtract the known total for the chosen

12

scores from the class total

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The sum of the scores of everyone in the class is

8k.

The sum of the scores in the collection of

12

is

12 \cdot 14 = 168.

This means that the sum of the scores of everyone not in the collection is

8k - 168.

There are also

k - 12

people not in the collection. Therefore, the average is

\dfrac{8k - 168}{k - 12}.

Thus, B is the correct answer.

Problem 5 in other years