Upstream Downstream Boat Math Problem

Problem Statement

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

Strategy

The boat problem usually has 3 components to it

speed of the river $S_{r}$
speed of the boat $S_{b}$
time to cover the distance (one way/round trip)

Downstream speed would be

S_{down} = S_{b} + S_{r}

And upstream speed would be

S_{up} = S_{b} - S_{r}

The time taken to complete the round trip would be

T = \frac{D}{S_{up}} + \frac{D}{S_{down}}

Condition given in the problem

If speed of the boat is doubled, time reduces by 75% (i.e its 25% of original time)

\frac{1}{4} T = \frac{D}{({2S}_{b} - S_{r})} + \frac{D}{({2S}_{b} + S_{r})}

Let's make T 100% in the equation

T = \frac{4D}{({2S}_{b} - S_{r})} + \frac{4D}{({2S}_{b} + S_{r})}

Finding the ratio now

\frac{D}{(S_{b} - S_{r})} + \frac{D}{(S_{b} + S_{r})} = \frac{4D}{({2S}_{b} - S_{r})} + \frac{4D}{({2S}_{b} + S_{r})}

The distance can be removed from both LHS and RHS

\frac{1}{(S_{b} - S_{r})} + \frac{1}{(S_{b} + S_{r})} = 4 (\frac{1}{({2S}_{b} - S_{r})} + \frac{1}{({2S}_{b} + S_{r})})

Let's try to reduce the equations.

\frac{{2S}_{b}}{({S_{b}}^{2} - {S_{r}}^{2})} = 4 (\frac{{4S}_{b}}{({4S}_{b}^{2} - {S_{r}}^{2})})

And reduce it even further

8 ({S_{b}}^{2} - {S_{r}}^{2}) = ({4S}_{b}^{2} - {S_{r}}^{2})

Moving the sides around we get

{8S}_{b}^{2} - {4S}_{b}^{2} = {8S}_{r}^{2} - {S_{r}}^{2}

The ratio of the squares would be.

\frac{{S_{b}}^{2}}{{S_{r}}^{2}} = \frac{7}{4}

Answer

\frac{S_{b}}{S_{r}} = \frac{\sqrt{7}}{2}