Full explanation
Let a, b and o be the numbers of apples (30c), bananas (40c) and oranges (50c) bought, so 30a + 40b + 50o = 200, with a, b and o all non-negative whole numbers.
Statement 1 — he cannot have exactly 3 apples: with a = 3, the apples cost 90c, leaving 110c for 40b + 50o = 110. Testing whole values of b: b = 1 needs o = 70/50 = 1.4; b = 2 needs o = 10/50 = 0.2; b = 0 needs o = 110/50 = 2.2. None of these give a whole number of oranges, so 3 apples is impossible — statement 1 is correct.
Statement 2 — he must have at least one banana: try b = 0. Then 30a + 50o = 200, and a = 0, o = 4 is a whole-number solution (4 oranges, no apples or bananas, costing exactly 200c). Since a valid basket with zero bananas exists, statement 2 is false.
Statement 3 — he has 4, 5 or 6 fruits in total: checking every whole-number solution of 30a + 40b + 50o = 200 gives exactly six baskets: (a,b,o) = (4,0,0) and (5,0,1), each 6 fruit; (0,5,0), (1,3,1) and (2,1,2), each 5 fruit; and (0,0,4), 4 fruit. Every basket totals 4, 5 or 6 fruit, so statement 3 is correct.
Statement 4 — if he has all three types, apples are fewer than bananas plus oranges combined: only two of the baskets above use all three fruit types: (1,3,1) — 1 apple against 3 + 1 = 4 bananas and oranges — and (2,1,2) — 2 apples against 1 + 2 = 3. In both, apples are fewer than the other two combined, so statement 4 is correct.
Statements 1, 3 and 4 hold and statement 2 fails, so the answer is C.
📚 Summary to learn
A word problem built around a linear equation with several unknowns (here, 30a + 40b + 50o = 200) is worked through by testing whether whole-number solutions exist for the unknowns under different assumptions — a statement claiming something is impossible is true only if no combination of non-negative whole numbers satisfies the equation under that assumption, and a statement claiming something is possible is true as soon as a single valid whole-number solution is found.