1. Write the radioactive decay law

For radioactive decay, $N = N_0 e^{-kt}$ where $k$ is the decay constant.

Also, $k = \frac{\ln 2}{t_{1/2}}$

So for substances $A$ and $B$:

$k_A = \frac{\ln 2}{15}, \qquad k_B = \frac{\ln 2}{5}$

2. Use the given initial concentrations

Let the initial concentration of $A$ be $a$. Then initial concentration of $B$ is $4a$

At time $t$, $[A] = a e^{-k_A t}$ $[B] = 4a e^{-k_B t}$

We need the time when their concentrations become equal: $a e^{-k_A t} = 4a e^{-k_B t}$

Cancel $a$: $e^{-k_A t} = 4 e^{-k_B t}$

Rearrange: $e^{(k_B-k_A)t} = 4$

3. Substitute the decay constants

$k_B-k_A = \frac{\ln 2}{5} - \frac{\ln 2}{15}$

Take LCM: $k_B-k_A = \ln 2\left(\frac{3-1}{15}\right)=\frac{2\ln 2}{15}$

Thus, $e^{\left(\frac{2\ln 2}{15}\right)t} = 4$

Since $4 = 2^2 = e^{2\ln 2}$ we get $\frac{2\ln 2}{15}t = 2\ln 2$

Cancelling $2\ln 2$: $t = 15 \text{ min}$

4. Final answer

$\boxed{15}$ min