AMC 10 Step by Step

2021 AMC Fall 10A Problem 10

P1-10~3 minPrint

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 50,20,20,5,50, 20, 20, 5, and 55 . 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−st-s ?

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Key insight

t is the plain average, 20. For s, each student reports their own class size, so the big class counts 50 times: s = 33.5.

Solution

1. Find tt. Each teacher reports the size of their own class, and each teacher is equally likely. So tt is the ordinary average of the five numbers:
t=50+20+20+5+55=1005=20.t = \frac{50 + 20 + 20 + 5 + 5}{5} = \frac{100}{5} = 20.
2. Find ss: what does each student say? Every student reports the size of their own class. List what all 100100 students say:
- The 5050 students in the big class each say 5050. Together: 50×50=250050 \times 50 = 2500.
- The 20+20=4020 + 20 = 40 students in the two middle classes each say 2020. Together: 40×20=80040 \times 20 = 800.
- The 5+5=105 + 5 = 10 students in the two small classes each say 55. Together: 10×5=5010 \times 5 = 50.
3. Average those 100100 answers.
s=2500+800+50100=3350100=33.5.s = \frac{2500 + 800 + 50}{100} = \frac{3350}{100} = 33.5.
4. Subtract. t−s=20−33.5=−13.5t - s = 20 - 33.5 = -13.5.

The answer is (B) −13.5\boxed{\textbf{(B)}\ {-}13.5}.

Why this works

When you average over teachers, each class counts once. When you average over students, each class counts once for every student in it, so the big class gets a much louder voice. That is why ss is bigger than tt, and you can tell before computing that t−st - s must be negative, leaving only (A) or (B). This is the reason students often feel their classes are bigger than the school's "average class size".

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

For tt, each of the 55 teachers is equally likely, so tt is the ordinary mean of the five class sizes:
t=50+20+20+5+55=1005=20.t = \frac{50 + 20 + 20 + 5 + 5}{5} = \frac{100}{5} = 20.

For ss, each of the 100100 students is equally likely, and a student in a class of size nn reports nn. There are 5050 students who report 5050, 4040 who report 2020, and 1010 who report 55:
s=50⋅50+20⋅20+20⋅20+5⋅5+5⋅5100=2500+800+50100=33.5.s = \frac{50 \cdot 50 + 20 \cdot 20 + 20 \cdot 20 + 5 \cdot 5 + 5 \cdot 5}{100} = \frac{2500 + 800 + 50}{100} = 33.5.

Therefore t−s=20−33.5=−13.5t - s = 20 - 33.5 = -13.5.

The answer is (B) −13.5\boxed{\textbf{(B)}\ {-}13.5}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.