Two substances A and B are present, such that $\left[A_0\right]=4\left[B_0\right]$ and the half life of A is 5 minutes and that of B is 15 minutes. If they start decaying at the same time following first order kinetics, how much time later will the concentration of both of them would be the same?

Amount of $A$ left in $n_1$ half lives $={\left(\frac{1}{2}\right)}^n\left[Ao\right]$

Amount of $B$ left $n_2$ half-lives $={\left(\frac{1}{2}\right)}^{n_2}\left[Bo\right]$

At the end, $\frac{\left[Ao\right]}{2^{n_1}}=\frac{\left[Bo\right]}{2^{n_2}}\Rightarrow \frac{4}{2^{n_1}}=\frac{1}{2^{n_2}}\{\left[A_0\right]=4\left[B_0\right]\}$

$\frac{2^{n_1}}{2^{n2}}=4\Rightarrow 2^{n_1-n_2}={\left(2\right)}^2$

$n_1-n_2=2\Rightarrow n_2=n_1-2$

(Let concentration of both become equal after $time\text{\hspace{0.17em}}t$ )

$\frac{n_1\times t_{\frac{1}{2}}\left(A\right)}{n_2\times t_{\frac{1}{2}}\left(B\right)}=1 \frac{n_1\times 5}{n_2\times 15}=1$

For $eq\left(1\right)\&\left(2\right)$

$n_1=3 n_2=1$

$t=3\times 5=15\min utes$