A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? _____ min.

We have half-life of A = 15 min, half-life of B = 5 min, and the initial concentration of B = 4 times that of A, i.e., $$[B]_0 = 4[A]_0$$.

For radioactive decay, the concentration at time $$t$$ is:

$$[A]_t = [A]_0 \left(\frac{1}{2}\right)^{t/15}$$

$$[B]_t = 4[A]_0 \left(\frac{1}{2}\right)^{t/5}$$

Setting $$[A]_t = [B]_t$$:

$$[A]_0 \left(\frac{1}{2}\right)^{t/15} = 4[A]_0 \left(\frac{1}{2}\right)^{t/5}$$

$$\left(\frac{1}{2}\right)^{t/15} = 4 \left(\frac{1}{2}\right)^{t/5} = \left(\frac{1}{2}\right)^{-2} \cdot \left(\frac{1}{2}\right)^{t/5} = \left(\frac{1}{2}\right)^{t/5 - 2}$$

Comparing exponents:

$$\frac{t}{15} = \frac{t}{5} - 2$$

$$\frac{t - 3t}{15} = -2$$

$$\frac{-2t}{15} = -2$$

Hence, $$t = 15$$ min. So, the answer is $$15$$ min.