Law of Radioactive Decay question from MHT CET 2024 (4 May, Shift 2)

Answer: (c)

1. In 120 min: A completes 4 half-lives, B completes 2 half-lives.
2. Decayed A $= N_0(1 - \frac{1}{16}) = \frac{15}{16}N_0$; decayed B $= N_0(1 - \frac{1}{4}) = \frac{12}{16}N_0$.
3. B : A $= 12 : 15 = 4 : 5$.
4. So the answer is (c) $4 : 5$.