Let the News, Features, and Sports counts be a, b, and g. There are two constraints: a + b + g = 30 editors, and 30a + 60b + 90g = 50 × 30 = 1,500 articles. Dividing the article equation by 30 gives a + 2b + 3g = 50.
Subtract the count equation from this to eliminate a: (a + 2b + 3g) − (a + b + g) = 50 − 30, so b + 2g = 20, giving b = 20 − 2g. Substituting back, a = 30 − b − g = 30 − (20 − 2g) − g = g + 10.
The reflex to minimize every other desk in order to maximize Sports is the trap here. Since a = g + 10, News grows as Sports grows and cannot be minimized; the only count free to shrink is Features, b = 20 − 2g. So the binding floor is Features ≥ 1. From 20 − 2g ≥ 1 we get 2g ≤ 19, so g ≤ 9.5, and since g is a whole number, g ≤ 9. At g = 9, Features = 2 and News = 19, and 19 + 2 + 9 = 30 with (30 × 19 + 60 × 2 + 90 × 9) ÷ 30 = 1,500 ÷ 30 = 50. So the greatest possible Sports count is 9, and the answer is D.
(A) 6, (B) 7, and (C) 8 are all feasible splits (News 16/17/18, Features 8/6/4), so any trial-and-stop approach lands on one of them; only pushing Features down to its floor of 2 reveals the true maximum. (E) 10 drops the at-least-one floor on Features (or assumes an equal 10-10-10 split, or rounds 9.5 up), which is the dominant trap. A two-desk alligation lever does not apply with three desks, and the choices cannot simply be plugged in without first deriving b = 20 − 2g and a = g + 10, so the system reduction plus the integer floor is unavoidable.