Competition · AMC preparation · step 4 of 4

AMC 10 · 2021B · #19

Grade 6 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

mean-median-mode-rangesystems-of-equationslinear-equations-two-vardecimal-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: mean-median-mode-rangesystems-of-equations

📏 Medium solution 💡 3 insights

Problem

S is a finite set of positive integers. Three averaging conditions are given: remove the greatest integer G and the average drops to 32; remove L (the least) as well and the average becomes 35; put G back (so L is the only one removed) and the average is 40. Also, G - L = 72. Find the average of all integers in the original set S.

Pick an answer.

(A)

36.2

(B)

36.4

(C)

36.6

(D)

36.8

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

Tool #13 (Algebra) — the four sentences are tailor-made to be turned into four equations in n, Σ, G, L. Once written, the system isn't messy: the two "average over n-1" lines differ only in whether G or L is removed, so subtracting them isolates G - L = 8(n - 1) and uses condition (4) to get n in one step. Tool #7 (Identify Subproblems) sequences the unknowns: first n, then L (from the n - 2 equation), then Σ (back into one of the n - 1 equations). Finally divide Σ by n for the answer. Each sub-step is just careful arithmetic — no need for complex algebra moves.

1STEP 1

Name the count and the sum

Let n = number of integers in S and Σ = total sum. Translate the four English sentences into four algebraic equations.

(1) Σ - G &= 32(n-1) ; (2) Σ - G - L &= 35(n-2) ; (3) Σ - L &= 40(n-1) ; (4) G - L &= 72

2STEP 2

Subtract to isolate the gap

Subtract (1) from (3): the sums cancel, leaving G - L = 8(n - 1).

(3) - (1): G - L = (40 - 32)(n - 1) = 8(n - 1)

3STEP 3

Solve for the count

Combine with (4): 72 = 8(n - 1), so n - 1 = 9 and n = 10.

72 = 8(n - 1) → n - 1 = 9 → n = 10

4STEP 4

Find the smallest number

Plug n = 10 into (1) and (2): Σ - G = 288 and Σ - G - L = 280, so L = 8.

Σ - G = 288, Σ - G - L = 280 → L = 8

5STEP 5

Find the total sum

Plug L = 8 into (3): Σ - 8 = 360, so Σ = 368.

Σ = 40 · 9 + L = 360 + 8 = 368

6STEP 6

Divide for the average

Average of S: $\frac{Σ}{n}$ = $\frac{368}{10}$ = 36.8, which is choice (D).

Σ/n = 368/10 = 36.8 → (D)

Answer

36.8

Verify with G: equation (4) gives G = L + 72 = 8 + 72 = 80. Check equation (1): Σ - G = 368 - 80 = 288 = 32 · 9 ✓. Check equation (2): Σ - G - L = 368 - 80 - 8 = 280 = 35 · 8 ✓. Check equation (3): Σ - L = 368 - 8 = 360 = 40 · 9 ✓. All four conditions hold, and the answer 36.8 sits squarely between 32 (lowest reported avg) and 40 (highest) — exactly the kind of number a balanced average should produce.

💡Key takeaway

This AMC 10 problem only needs Grade 6 "sum = average × count" equations you already know! Subtract two n - 1 averages to get G - L = 8(n-1) = 72, so n = 10. Then Σ = 368 and the answer is $\frac{368}{10}$ = 36.8.

Name the count and the sum
Subtract to isolate the gap
Solve for the count
Find the smallest number
Find the total sum
Divide for the average

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