AMC 10 · 2023 · #18

Grade 7 algebralogic
weighted-averagemean-median-mode-rangeextreme-principlelogical-deduction convert-to-algebraextreme-principlelogical-deduction ↑ Prerequisites: mean-median-mode-rangeweighted-average
📏 Long solution 💡 4 insights
Problem
Two students took quizzes over two semesters, and the quiz counts need not match between students or between semesters. Three facts are given: the first-semester gap, one student's rise between semesters, and the second-semester gap. Each student's year average is the average over all her quizzes. Of five statements about the year averages, exactly one cannot be arranged. Find it.

Pick an answer.

(A)
Yolanda's quiz average for the academic year was } 22 \text{ points higher than Zelda's.
(B)
Zelda's quiz average for the academic year was higher than Yolanda's.
(C)
Yolanda's quiz average for the academic year was } 3 \text{ points higher than Zelda's.
(D)
Zelda's quiz average for the academic year equaled Yolanda's.
(E)
If Zelda had scored } 3 \text{ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.
How to solve
Strategy Extreme Principle

Every statement is about one number: D = Y_avg - Z_avg. So the real question is which values D can take. Tool #13 (Convert to Algebra) turns the three sentences into three equations and exposes the hidden symmetry — Zelda improved by 18 too. Tool #4 (Introduce a Variable) names the four quiz counts, which is where the whole difficulty lives: the year average is a WEIGHTED mean, and the two students may weight their semesters differently. That collapses D into D = 3 + 18(w_Y - w_Z), where each w is the share of that student's quizzes falling in semester two. Tool #14 (Extreme Principle) then does the decisive work: push w_Y and w_Z to their boundaries to find the largest and smallest D can ever be. Tool #6 (Guess and Check) supplies an explicit set of quiz counts and scores witnessing each statement that IS achievable, and Tool #3 (Eliminate Possibilities) crosses those four off, leaving the one that lies outside the range.

1STEP 1

Turn the sentences into equations

The three sentences become three equations.

Y₁ = Z₁ + 3, Y₂ = Y₁ + 18, Y₂ = Z₂ + 3 ⟹ Z₂ = Z₁ + 18
2STEP 2

The year average is weighted

The year average is a weighted mean.

Y_avg = (n_Y1Y₁ + n_Y2Y₂)/(n_Y1 + n_Y2), Z_avg = (n_Z1Z₁ + n_Z2Z₂)/(n_Z1 + n_Z2)
3STEP 3

Base plus a share

Write each as a base plus a share.

Y_avg = Y₁ + 18 w_Y, Z_avg = Z₁ + 18 w_Z, w_Y = n_Y2/(n_Y1+n_Y2), w_Z = n_Z2/(n_Z1+n_Z2)
4STEP 4

One formula for the gap

The gap tidies into one formula.

D = Y_avg - Z_avg = (Y₁ - Z₁) + 18(w_Y - w_Z) = 3 + 18(w_Y - w_Z)
5STEP 5

Push the shares to their extremes

Pushing the shares gives a range.

0 ≤ w_Y, w_Z ≤ 1 ⟹ -1 ≤ w_Y - w_Z ≤ 1 ⟹ 3 - 18 ≤ D ≤ 3 + 18, so -15 ≤ D ≤ 21
6STEP 6

Build a witness for each

Build an actual example for each reachable claim.

(B) 66 < 75. (C) 72 - 69 = 3. (E) (63+63+81+81)/4 = 72 = Y_avg. (D) 69 = 69.
7STEP 7

Keep the impossible one

The only claim beyond the range is a twenty-two point gap.

D_max = 21 < 22 ⟹ (A) is unreachable → (A)
Answer
Yolanda's quiz average for the academic year was 22 points higher than Zelda's.
The bound passes a common-sense test. Yolanda is ahead by 3 in each semester separately, so if the two girls took quizzes on the same schedule the yearly gap would be exactly 3 — and 3 sits comfortably inside [-15, 21], matching choice (C). The extra room comes only from mismatched schedules, and the most a mismatch can be worth is the full 18-point raise, giving the ceiling 3 + 18 = 21 and the floor 3 - 18 = -15. The number 22 in (A) is deliberately set one point above that ceiling, which is exactly the kind of near-miss that rewards computing the bound instead of guessing. A brute-force sweep over all quiz counts from 1 to 40 in each of the four slots confirms it: the largest gap found was 825/41 ≈ 20.12 (Yolanda 1 then 40 quizzes, Zelda 40 then 1), climbing toward 21 but never reaching it. Each of the four witnesses was also checked by direct division rather than by formula, and all four came out as claimed.
💡Key takeaway

A year average is a weighted average, so the same 18-point improvement is worth more to whoever took more quizzes afterward — and since that lever is worth at most 18 points on top of Yolanda's steady 3-point edge, her yearly lead can never pass 21.

  • Turn the three sentences into equations
  • The year average is a weighted mean
  • Rewrite each average as base plus a share of 18
  • One formula for the yearly gap
  • Push the shares to their extremes
  • Build a witness for each achievable claim
  • Cross off four, keep the impossible one