The marks of 70 candidates are grouped as 0–10: 5, 10–20: f₁, 20–30: 20, 30–40: 15, 40–50: f₂, and the median mark is 28. Find f₁ and f₂.

Two facts are needed, because one unknown cannot be pinned down by the total alone. First, the frequencies add to 70, so 5 + f₁ + 20 + 15 + f₂ = 70 and f₁ + f₂ = 30. Second, the median 28 lies in the class 20–30, where l = 20, h = 10, f = 20 and the preceding cumulative frequency is cf = 5 + f₁, with n/2 = 35. The median formula gives 28 = 20 + ((35 − (5 + f₁))/20) × 10, so 8 = (30 − f₁)/2, giving f₁ = 14 and hence f₂ = 16. Check: the cumulative frequencies run 5, 19, 39, 54, 70, the class 20–30 is indeed the median class, and 20 + ((35 − 19)/20) × 10 = 28. Each wrong option satisfies f₁ + f₂ = 30 and so passes the total test, but fails the median test — they give medians of 27, 27.5 and 29 respectively. That is exactly why BOTH conditions must be used.