If the linear momentum is increased by 50 \% , then kinetic energy...

Kinetic energy of a body, $K=\frac{p^2}{2 m}$ where $p$ is the momentum of the body and $m$ is the mass of the body $p^{\prime}=p+\frac{50}{100} p=\frac{3}{2} p$ Assuming $m$ remains constant $\therefore \frac{K^{\prime}}{K}=$\left(\frac{p^{\prime}}{p}\right)^2=\frac{9}{4} \%$increase in kinetic energy$=\frac{K^{\prime}-K}{K} \times 100 \% =\left(\frac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\frac{9}{4}-1\right) \times 100 \%=125 \%