AMC 10 · 2013 · #16

Grade 7 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
mean-median-mode-rangeweighted-averageoptimization extreme-principleconvert-to-algebra ↑ Prerequisites: mean-median-mode-range
📏 Long solution 💡 3 insights
Problem
Several averages are known but one pile's own average never is. Find the largest whole-number average of the last combination.

Pick an answer.

(A)
55
(B)
56
(C)
57
(D)
58
(E)
59
How to solve
Strategy Extreme Principle

The word "greatest" means the job has two halves that both need doing: show the target mean can never reach some ceiling, and then build an actual pile arrangement that lands on the biggest integer under it. Name the three rock counts and pile C's total weight, turn each given mean into an equation about totals, and squeeze the target mean into the single shape 44 + (constant)×b/(b+c). That shape makes the ceiling obvious, because b/(b+c) can crowd up toward 1 but never touch it. Then solve backwards for the rock counts that hit the integer just below the ceiling.

1STEP 1

Trade every mean for a total

Each average becomes a total.

mean = (total weight)/(rock count); W_A = 40a, W_B = 50b, W_C = w
2STEP 2

Piles A and B fix a ratio

The first pair fixes a ratio.

(40a+50b)/(a+b) = 43 ⟹ 40a+50b = 43a+43b ⟹ 3a = 7b ⟹ a = 7/3b
3STEP 3

Piles A and C fix C's total weight

The second pair fixes the unknown total.

(40a + w)/(a+c) = 44 ⟹ 40a + w = 44a + 44c ⟹ w = 4a + 44c
4STEP 4

Rewrite the target in one clean shape

The target rewrites as a constant plus a fraction.

M = (50b + w)/(b+c) = (178/3b + 44c)/(b+c) = (44(b+c) + 46/3b)/(b+c) = 44 + 46/3·b/(b+c)
5STEP 5

Read off the ceiling

That fraction stays below one, giving a ceiling.

b ≥ 1, c ≥ 1 ⟹ 0 ＜ b/(b+c) ＜ 1 ⟹ 44 ＜ M ＜ 178/3 = 59 1/3
6STEP 6

Build an arrangement that hits 59

A real arrangement reaches 59, choice (E).

b/(b+c) = 45/46 ⇔ b = 45c; (a,b,c) = (105,45,1), w = 464; (2250+464)/46 = 59
Answer
59
The explicit piles (a,b,c) = (105,45,1) with weights 4200, 2250, 464 satisfy every stated condition: 4200/105 = 40, 2250/45 = 50, (4200+2250)/150 = 43, and (4200+464)/106 = 44, while (2250+464)/46 = 59. So 59 is genuinely reachable. On the other side, the exact ceiling is 178/3 = 59 1/3, never attained because that needs c = 0; this explains the answer list, which stops at 59 precisely because the next integer is just past the ceiling. A quick feel check: the target mean must sit above 50 since pile C is being asked to be very heavy, and it should stay near 59 rather than run away, since pile B's own 50-pound rocks always drag the combination down.
💡Key takeaway

Turn every average into a total, squeeze the answer into "44 plus a leftover" so only one ratio is free, then prove the biggest value twice: nothing goes higher, and this exact pile does reach it.

  • Trade every mean for a total
  • Piles A and B fix a ratio
  • Piles A and C fix C's total weight
  • Rewrite the target in one clean shape
  • Read off the ceiling
  • Build an arrangement that hits 59

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