AMC 8 · 2000 · #23

Grade 6 arithmetic
Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region
mean-median-mode-rangeset-partitionfraction-arithmeticmulti-digit-arithmetic identify-subproblemscomplementary-counting ↑ Prerequisites: mean-median-mode-rangefraction-arithmetic
📏 Short solution 💡 3 insights
Problem
Seven numbers sit in a row. The average of the first four is 5, and the average of the last four is 8. The average of all seven is 6 47\frac{4}{7}. The first-four group and the last-four group overlap on one number (the same number is in both). Find that overlapping number.

Pick an answer.

(A)
$5\frac{3}{7}$
(B)
6
(C)
$6\frac{4}{7}$
(D)
7
(E)
$7\frac{3}{7}$

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Find an Invariant

We never need the seven individual numbers — only their sums. Tool #11 (Find an Invariant) says: the grand total of all seven numbers is the same no matter how you slice them. Tool #9 (Solve an Easier Problem) is the matching move — instead of seven unknowns, turn each average into a sum and reason about three totals: first-four sum, last-four sum, and all-seven sum. Adding the first-four sum and the last-four sum double-counts exactly the overlapping number, so (first-four sum) + (last-four sum) - (all-seven sum) = overlap. No variables, no algebra.

1STEP 1

Turn each average into a sum with sum = average × count — apply it to the first four, the last four, and all seven.

first-4 sum = 4 × 5 = 20, last-4 sum = 4 × 8 = 32, all-7 sum = 7 × 6 47\frac{4}{7} = 7 · 6 + 4 = 46
2STEP 2

Add the two group sums to 52 — the shared number sits in both fours, so it gets counted twice.

first-4 sum + last-4 sum = 20 + 32 = 52 = (all-7 sum) + (overlap)
3STEP 3

Subtract the all-seven total from the combined sum: 52 - 46 = 6 is the double-counted shared number, choice (B).

overlap = 52 - 46 = 6 → (B)
Answer
6
The overlap 6 should be a plausible value for a number that belongs to both groups. The first-four average is 5, so numbers there hover near 5; the last-four average is 8, so numbers there hover near 8. A shared number sitting between 5 and 8 fits — and 6 lies in that range. Also, 7 × 6 47\frac{4}{7} = 46 matches: 20 + 32 - 6 = 46, exactly the all-seven sum. Choices (A) 5 37\frac{3}{7} and (E) 7 37\frac{3}{7} would also balance the arithmetic only if the totals were different, so we can rule them out by the calculation above.
💡Key takeaway

Convert the averages into sums: 20, 32, 46. When you add 20 and 32, the shared number gets counted twice — so 52 - 46 = 6 is exactly that shared number. Answer (B).

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