2021 AMC 10A Spring Problem 5

Below is the video solution and professionally curated solution for Problem 5 of the 2021 AMC 10A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Spring solutions, or check the answer key.

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Concepts:meanalgebraic manipulation

Difficulty rating: 900

5.

The quiz scores of a class with k>12k > 12 students have a mean of 8.8. The mean of a collection of 1212 of these quiz scores is 14.14. What is the mean of the remaining quiz scores in terms of k?k?

148k12\dfrac{14-8}{k-12}

8k168k12\dfrac{8k-168}{k-12}

14128k\dfrac{14}{12} - \dfrac{8}{k}

14(k12)k2\dfrac{14(k-12)}{k^2}

14(k12)8k\dfrac{14(k-12)}{8k}

Video solution:
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Written solution:

The sum of the scores of everyone in the class is 8k.8k. The sum of the scores in the collection of 1212 is 1214=168.12 \cdot 14 = 168.

This means that the sum of the scores of everyone not in the collection is 8k168.8k - 168. There are also k12k - 12 people not in the collection. Therefore, the average is 8k168k12. \dfrac{8k - 168}{k - 12}.

Thus, B is the correct answer.

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