2021 AMC 10A Fall Problem 10

Attempt Problem 10 of the 2021 AMC 10A Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Fall solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

A school has 100100 students and 55 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are 50,50, 20,20, 20,20, 5,5, and 5.5. Let tt be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let ss be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is t−s?t-s?
−18.5-18.5
−13.5-13.5
00
13.513.5
18.518.5
Answer: B
Concepts:expected valuemean
Small hint:
The teacher average weights each class once
Big hint:
The student average weights a class by its enrollment
Solution:
Recall the expected-value formula E[X]=∑xx Pr⁡(X=x). \mathbb E[X]=\sum_x x\,\Pr(X=x).
Therefore, t=15(50+20+20+5+5) t = \dfrac{1}{5} (50 + 20 + 20 + 5 + 5) =15⋅100=20 = \dfrac{1}{5} \cdot 100 = 20 and s=50⋅50100+20⋅20100 s = 50 \cdot \dfrac{50}{100} + 20 \cdot \dfrac{20}{100} +20⋅20100+5⋅5100+5⋅5100 + 20 \cdot \dfrac{20}{100} + 5 \cdot \dfrac{5}{100} + 5 \cdot \dfrac{5}{100} =25+4+4+0.25+0.25=33.5 \begin{aligned} &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5 \end{aligned}
t−s=−13.5.t - s = -13.5.
Thus, B is the correct answer.
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