A and B are two substances undergoing radioactive decay in a...

For a first order reaction:
$⇒ [ A{]}_{t} = [ A{]}_{0}{e}^{- kt}$

For $A$ : Let $[ A{]}_{t}$ be $y$ and $[ A{]}_{0}$ be x
For a first order reaction:
$∴ k = \dfrac{\ln 2}{{t}_{1/2}} = \dfrac{\ln 2}{15min}$

$⇒ y = {xe}^{- kt}$

$= {xe}^{- \left(\dfrac{\ln 2}{15}\right)t}$ ....(1)

Similarly For B:
$[ B{]}_{t} = [ B{]}_{0}{e}^{- kt}$

Let $[ B{]}_{t} = y; [ B{]}_{0} = 4x; k = \dfrac{\ln 2}{{t}_{1/2}} = \dfrac{\ln 2}{5min}$

$⇒ y = 4{xe}^{- \left(\dfrac{\ln 2}{5}\right)t}$ ....(2)
Equating equation(1) and (2) we get:

$⇒ {xe}^{- \left(\dfrac{\ln 2}{15}\right)t} = 4{xe}^{- \left(\dfrac{\ln 2}{5}\right)t}$

$⇒ t \times \left[\dfrac{\ln 2}{5} - \dfrac{\ln 2}{15}\right] = \ln 4$

$⇒ t \times \ln 2\left[\dfrac{1}{5} - \dfrac{1}{15}\right] = 2\ln 2$

$⇒ {e}^{t\left(\dfrac{\ln 2}{5} - \dfrac{\ln 2}{15}\right)} = 4$

$⇒ t = 15min$