Q.If 75% of a first order reaction was completed in 60 min, 50% of the same reaction under the same conditions would be completed in :

Using $t=\dfrac{2.303}{k}\log\dfrac{a}{a-x}$, 75% completion needs $\log 4$ and 50% completion needs $\log 2$; since $\log 4=2\log 2$, the 75%-completion time is exactly twice the 50%-completion time, giving $t_{50\%}=30$ min.

For a first-order reaction, $t=\dfrac{2.303}{k}\log\dfrac{a}{a-x}$. At 75% completion, $x=0.75a$, so $a-x=0.25a$ and $\dfrac{a}{a-x}=4$; thus $60=\dfrac{2.303}{k}\log 4$. At 50% completion (this is simply $t_{1/2}$), $x=0.5a$, so $a-x=0.5a$ and $\dfrac{a}{a-x}=2$; thus $t_{50\%}=\dfrac{2.303}{k}\log 2$. Since $\log 4=\log(2^2)=2\log 2$, the 75%-completion time is exactly double the 50%-completion time: $60=2\times t_{50\%}$, so $t_{50\%}=30$ …