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Five distinct positive integers have an average (arithmetic mean) of 10. Is the largest of the five greater than 18?
(1) The smallest of the five integers is 2.
(2) The median (third-largest) of the five integers is 9.
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Correct answer
E
Five distinct positive integers averaging 10 sum to 50; we ask whether the maximum exceeds 18. This is the mean + count + per-element floor family.
Direction check first: to make the maximum large, push the other four as small as the distinctness rule allows; to make it small, spread the total. Both directions are tested against 18.
Statement (1): smallest is 2. NO-witness 2, 8, 9, 13, 18 (sum 50, largest 18, NO); yes-witness 2, 3, 4, 5, 36 (sum 50, largest 36, yes). Straddles. Not sufficient.
Statement (2): median is 9, so two values are below 9 and two above. NO-witness 4, 8, 9, 11, 18 (sum 50, largest 18, NO); yes-witness 1, 2, 9, 18, 20 (sum 50, largest 20, yes). Straddles. Not sufficient.
Together: smallest 2 AND median 9. The two values below 9 are 2 and one of {3,..., 8}; the two above 9 sum to 50 − (2 + a + 9) = 39 − a, where a is the second-smallest. NO-witness 2, 8, 9, 13, 18 (median 9, smallest 2, sum 50, largest 18, NO). yes-witness 2, 3, 9, 16, 20 (median 9, smallest 2, sum 50, largest 20, yes). Both satisfy both statements, and the largest straddles 18. Not sufficient even together.
The reflex plus the conjunction-tightens illusion (combined-sufficient illusion). Solvers think "smallest pinned and median pinned must force the max," choosing C. But two free values above the median can be split many ways (13 and 18, or 16 and 20), so the max is not pinned. The finding is the order-statistic bound: fixing the smallest and median still leaves the top two with one degree of freedom.
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