AMC 10 · 2021 · #7

Grade 7 probability
expected-valueweighted-averagemean-median-mode-rangeprobability-basic identify-subproblemsformula-substitution ↑ Prerequisites: expected-value
📏 Medium solution 💡 2 insights
Problem
A school splits 100 students among 5 classes, one class per teacher, with enrolments 50, 20, 20, 5, and 5. Pick a teacher at random and write down that class's size; the average is the first value. Pick a student at random and write down that class's size, counting the student too; the average is the second value. Find the first value minus the second.

Pick an answer.

(A)
${-}18.5$
(B)
${-}13.5$
(C)
0
(D)
13.5
(E)
18.5
How to solve
Strategy Organize Information in More Ways

There is only one set of data here — the five enrollments 50, 20, 20, 5, 5 — and the entire problem is that t and s tally it two different ways. That makes Tool #15 (Organize Information in More Ways) the engine: build the teacher list, then rebuild the very same classes as a student list and watch the numbers change. Tool #16 (Change Focus) names the switch precisely — stop looking at the 5 teachers and look at the 100 students, which is what turns a plain average into a weighted one. Tool #8 (Analyze the Units) keeps the two divisions straight, since t divides by 5 teachers while s divides by 100 students. Tool #14 (Extreme Principle) settles the sign of the answer before any arithmetic, by testing the all-equal case and a lopsided case. Tool #3 (Eliminate Possibilities) then cashes that sign in against the five choices.

1STEP 1

Tally the classes by teacher

By teacher it is a plain average.

t = (50 + 20 + 20 + 5 + 5)/5 = 100/5 = 20
2STEP 2

Re-tally the same classes by student

By student, big classes weigh more.

P(50) = 50/100 = 1/2, P(20) = 20/100 = 1/5 for each 20-class, P(5) = 5/100 = 1/20 for each 5-class
3STEP 3

Average over the hundred students

Take the average over the hundred students.

s = (50 · 50 + 20 · 20 + 20 · 20 + 5 · 5 + 5 · 5)/100 = 3350/100 = 33.5
4STEP 4

Decide the sign before finishing

The sign can be predicted in advance.

all classes equal → t = s; classes uneven → s > t → t - s < 0
5STEP 5

Subtract and match a choice

Subtracting gives negative thirteen point five.

t - s = 20 - 33.5 = -13.5
Answer
-13.5
Three checks. First, a range check: s is an average of numbers that are all between 5 and 50, so s must land in that range — and 33.5 does. It should also sit above the halfway mark, because half of all students are in the 50-class and that class alone contributes 50 · 1/2 = 25 to s before any other class is added; 33.5 clears that. Second, a structural check: s can be written as (sum of the squared class sizes)/(sum of the class sizes), and since the sum of the class sizes is 100 = 5t, a short rearrangement gives s = t + (variance of the five class sizes)/t. The five sizes have mean 20 and deviations 30, 0, 0, -15, -15, so the variance is (900 + 0 + 0 + 225 + 225)/5 = 1350/5 = 270, and 270/20 = 13.5. That reproduces the gap exactly, and it also explains why the gap can never be negative: a variance is never negative. Third, a units check: both t and s are counts of students per class, so subtracting them is legitimate and the answer is measured in students.
💡Key takeaway

Asking the teachers and asking the students are two different tallies of the same five classes: a class of n students is counted once by its teacher but n times by its students, so the student-side average is the bigger one — here 20 against 33.5, a gap of 13.5.

  • Tally the classes by teacher
  • Re-tally the same classes by student
  • Average over the hundred students
  • Decide the sign before finishing
  • Subtract and match a choice