From equation of radioactivity we know
$
A_{t}=A_{0} e^{-\lambda t}
$
Where $A _{ t }=$ Amount of material at time ' $t$ '
$A _{0}=$ Amount of substance at $t =0$
$\lambda=$ decay constant
$t =$ time.
Taking log we have
$\ln \left[\frac{A_{0}}{A_{t}}\right]=\lambda t$
For half life $A_{t}=\frac{A_{0}}{2}$.
$\Rightarrow \ln 2=\lambda t_{1 / 2}$ (A) $t_{1 / 2}=$ half life
From given condition.
$\ln \left[\frac{700}{500}\right]=\lambda(30 min )$
(A) $-(B)$
$\frac{\ln 2}{\ln [7 / 5]}=\frac{ t _{1 / 2}}{30}$
$\Rightarrow t _{1 / 2}=\frac{\ln 2 \times 30}{\ln [7 / 5]} \Rightarrow t _{1 / 2}=\frac{0.693 \times 30}{0.336}$
$\Rightarrow t _{1 / 2}=61.8 min$