Problem 17 · 1989 AJHSME
Hard
Algebra & Patterns
bound-the-average
The number N is between 9 and 17. The average of 6, 10, and N could be
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Answer: B — 10.
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Hint 1 of 3
The word 'could be' is the clue: N isn't fixed, so the average isn't either — it lives in a range. Find the two ends of that range by trying N at its smallest and largest.
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Hint 2 of 3
When a quantity slides between bounds, find the smallest and largest possible results; the answer must land inside that window.
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Hint 3 of 3
Average = (6 + 10 + N)⁄3 = (16 + N)⁄3. As N creeps from just above 9 to just below 17, watch where this lands — then see which single choice fits.
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Approach: bound the average between its extremes
- Write the average as (6 + 10 + N)⁄3 = (16 + N)⁄3. Since the average grows steadily as N grows, its smallest and largest values come from N's endpoints.
- Smallest: N just above 9 gives just above 25⁄3 ≈ 8.3. Largest: N just below 17 gives just below 33⁄3 = 11. So the average can be anything between roughly 8.3 and 11.
- Only 10 sits inside that window (it happens at N = 14). The why: a quantity built from a sliding input takes every value between its two extremes, so 'could be' questions are really 'is it inside the range?' questions.
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