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A software company surveyed 200 of its customers, every one of whom uses at least one of three tools: the editor, the debugger, and the profiler. Of the 200, exactly 150 use the editor, 130 use the debugger, and 110 use the profiler. What is the greatest possible number of these customers who use all three tools?
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Correct answer
A
Count total tool-memberships: 150 + 130 + 110 = 390 memberships among 200 customers, an excess of 190 over one-each. Since everyone uses at least one tool, each customer falls in exactly 1, 2, or 3 tools. If e2 use exactly two and e3 use exactly three, then e2 + 2(e3) = 190. To maximize e3, push e2 to 0, giving 2(e3) = 190, so e3 = 95.
Another view uses deficits: 50 customers lack the editor, 70 lack the debugger, and 90 lack the profiler, for 210 missing memberships. Anyone using all three has no deficits, so the other 200 − x customers must absorb all 210, and each can be missing at most two tools. Requiring 2(200 − x) ≥ 210 gives 200 − x ≥ 105, so x ≤ 95.
And 95 is achievable: 95 customers in all three, 55 only the editor, 35 only the debugger, 15 only the profiler. Then editor = 150, debugger = 130, profiler = 110, total = 200. Trying x = 96 fails, since the remaining 104 customers could absorb only 2 × 104 = 208 deficits, short of 210. So 95 is the greatest possible. So the answer is A.
C (110) is the reflex min(150, 130, 110), which ignores that the 200-customer total caps the overlap below the smallest group. D (150) caps at the largest group and never tests the total. B (105) is the count of non-triple customers, from solving the inequality but forgetting to recover x. E (190) is the raw membership excess, reported before splitting it into the exactly-two and exactly-three contributions.
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