AMC 10 · 2010 · #12

Grade 6 algebra
Archetype Overlap / Complement Counting └ Sub-archetype Two-Attribute Table: Fill or Optimize a Cell
optimizationpercentageset-partition extreme-principle ↑ Prerequisites: percentage
📏 Medium solution 💡 3 insights
Problem
At the start of the school year, 50% of a class said "Yes" to loving math and 50% said "No". At the end of the year, 70% said "Yes" and 30% said "No". Counting everyone whose answer differed between the two times, call that share x%. Find the difference between the maximum and minimum possible values of x.

Pick an answer.

(A)
0
(B)
20
(C)
40
(D)
60
(E)
80

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The question asks for the biggest and smallest a quantity can be, which is the signature of Tool #14 (Extreme Principle): solve by finding the boundary cases. Tool #15 (Organize Information in More Ways) sorts every student into a four-way start-to-end table so nobody is double counted. Tool #4 (Introduce a Variable) names the one free choice — how many students flip from "Yes" to "No" — turning the switch count into a single expression whose two ends give the answer.

1STEP 1

Sort students into four groups

Sort everyone by start-then-end answer into Yes→Yes, Yes→No, No→Yes, No→No; the two flip groups together make up x%.

x = (Yes→No) + (No→Yes)
2STEP 2

Name the flips and link them

Let s be the Yes→No percent. "Yes" must gain 20, so No→Yes minus s equals 20, forcing No→Yes = s + 20.

(No→Yes) - s = 20 → No→Yes = s + 20
3STEP 3

Write the switch count as one expression

Add the two flip groups and substitute No→Yes = s + 20, so the switchers come to x = 2s + 20.

x = s + (s + 20) = 2s + 20
4STEP 4

Push s to its extremes

Nobody is forced to flip Yes→No, so s starts at 0; only 30% end on "No", so s tops out at 30 — giving 0 ≤ s ≤ 30.

0 ≤ s ≤ 30
5STEP 5

Compare the extremes

s = 0 gives x = 20 and s = 30 gives x = 80, so the gap between them is 60.

x_min = 2(0)+20 = 20, x_max = 2(30)+20 = 80, 80 - 20 = 60 → (D)
Answer
60
Test both extreme cases against every total. Minimum (s = 0): Yes→No = 0, No→Yes = 20, Yes→Yes = 50, No→No = 30. Start Yes = 50+0 = 50, end Yes = 50+20 = 70, end No = 0+30 = 30, and changers = 0+20 = 20. Maximum (s = 30): Yes→No = 30, No→Yes = 50, Yes→Yes = 20, No→No = 0. Start Yes = 20+30 = 50, end Yes = 20+50 = 70, end No = 30+0 = 30, and changers = 30+50 = 80. Both cases satisfy all four totals, and 80 - 20 = 60, confirming (D).
💡Key takeaway

Sort everyone into who-switched groups, notice the Yes side only moves by the net in-minus-out, then push the free flip count to its lowest and highest to bracket the answer.

  • Sort students into four groups
  • Name the flips and link them
  • Write the switch count as one expression
  • Push s to its extremes
  • Compare the extremes

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