Upstream Downstream Boat Math Problem
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# 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

${S}_{\mathrm{down}}={S}_{b}+{S}_{r}$

And upstream speed would be

${S}_{\mathrm{up}}={S}_{b}-{S}_{r}$

The time taken to complete the round trip would be

$T=\frac{D}{{S}_{\mathrm{up}}}+\frac{D}{{S}_{\mathrm{down}}}$

### Condition given in the problem

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

$\frac{1}{4}T=\frac{D}{\left({\mathrm{2S}}_{b},-,{S}_{r}\right)}+\frac{D}{\left({\mathrm{2S}}_{b},+,{S}_{r}\right)}$

Let's make T 100% in the equation

$T=\frac{\mathrm{4D}}{\left({\mathrm{2S}}_{b},-,{S}_{r}\right)}+\frac{\mathrm{4D}}{\left({\mathrm{2S}}_{b},+,{S}_{r}\right)}$

### Finding the ratio now

$\frac{D}{\left({S}_{b},-,{S}_{r}\right)}+\frac{D}{\left({S}_{b},+,{S}_{r}\right)}=\frac{\mathrm{4D}}{\left({\mathrm{2S}}_{b},-,{S}_{r}\right)}+\frac{\mathrm{4D}}{\left({\mathrm{2S}}_{b},+,{S}_{r}\right)}$

The distance can be removed from both LHS and RHS

$\frac{1}{\left({S}_{b},-,{S}_{r}\right)}+\frac{1}{\left({S}_{b},+,{S}_{r}\right)}=4\left(\frac{1}{\left({\mathrm{2S}}_{b},-,{S}_{r}\right)},+,\frac{1}{\left({\mathrm{2S}}_{b},+,{S}_{r}\right)}\right)$

Let's try to reduce the equations.

$\frac{{\mathrm{2S}}_{b}}{\left({{S}_{b}}^{2},-,{{S}_{r}}^{2}\right)}=4\left(\frac{{\mathrm{4S}}_{b}}{\left({{\mathrm{4S}}_{b}}^{2},-,{{S}_{r}}^{2}\right)}\right)$

And reduce it even further

$8\left({{S}_{b}}^{2},-,{{S}_{r}}^{2}\right)=\left({{\mathrm{4S}}_{b}}^{2},-,{{S}_{r}}^{2}\right)$

Moving the sides around we get

${{\mathrm{8S}}_{b}}^{2}-{{\mathrm{4S}}_{b}}^{2}={{\mathrm{8S}}_{r}}^{2}-{{S}_{r}}^{2}$

The ratio of the squares would be.

$\frac{{{S}_{b}}^{2}}{{{S}_{r}}^{2}}=\frac{7}{4}$

$\frac{{S}_{b}}{{S}_{r}}=\frac{\sqrt{7}}{2}$