A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

The correct option is B.

Let the speed of the motor boat in still water be $v$ and the speed of the river be $r$.

Let the distance of the office from his home be $d$.

The original round-trip travel time is:
$T_1=\frac{d}{v+r}+\frac{d}{v-r}=\frac{2vd}{v^2-r^2}$

When the speed of the motor boat is doubled, the new speed is $2v$. The new round-trip travel time is:
$T_2=\frac{d}{2v+r}+\frac{d}{2v-r}=\frac{4vd}{4v^2-r^2}$

Since the total travel time gets reduced by 75%:
$T_2=0.25T_1=\frac{1}{4}$

Equating the expressions:
$\frac{4vd}{4v^2-r^2}=\frac{1}{4}$

Divide both sides by $2vd$:
$\frac{2}{4v^2-r^2}=\frac{1}{2\left(v^2-r^2\right)}$

Cross-multiplying gives:
$4\left(v^2-r^2\right)=4v^2-r^2$
$4v^2-4r^2=4v^2-r^2$ (Wait, let's write correct math)
Let's solve for ratio $x=\frac{v}{r}$:
$8\left(v^2-r^2\right)=4v^2-r^2$
$8v^2-8r^2=4v^2-r^2$
$4v^2=7r^2$
$\frac{v^2}{r^2}=\frac{7}{4}\Rightarrow \frac{v}{r}=\frac{\sqrt{7}}{2}$

Thus, the ratio is $\sqrt{7}:2$.