A motorboat covers a given distance in \(6 \mathrm{~h}\) moving downstream on a river. It covers the same distance in \(10 \mathrm{~h}\) moving upstream. The time it takes to cover the same distance in still water is

If \(\mathrm{v}_{\mathrm{w}}\) be the velocity of water and \(\mathrm{v}_{\mathrm{b}}\) be the velocity of motorboat in still water.
The distance covered by motorboat in moving downstream in \(6 \mathrm{~h}\) is
\(\mathrm{x}=\left(\mathrm{v}_{\mathrm{b}}+\mathrm{v}_{\mathrm{w}}\right) \times 6 \quad \text{...(i)}\)
Same distance covered by motorboat in moving upstream in \(10 \mathrm{~h}\) is
\(\mathrm{x}=\left(\mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{w}}\right) \times 10 \quad \text{...(ii)}\)
From Eqs. (i) and (ii), we have
\(\left(\mathrm{v}_{\mathrm{b}}+\mathrm{v}_{\mathrm{w}}\right) \times 6=\left(\mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{w}}\right) \times 10 \)
\(\mathrm{v}_{\mathrm{w}}=\frac{\mathrm{v}_{\mathrm{b}}}{4} \)
\(\therefore \mathrm{x}=\left(\mathrm{v}_{\mathrm{b}}+\mathrm{v}_{\mathrm{w}}\right) \times 6=7.5 \mathrm{v}_{\mathrm{b}}\)
Time taken by the motorboat to cover the same distance in still water is
\(t=\frac{x}{v_{b}}=\frac{7.5 v_{b}}{v_{b}}=7.5 \mathrm{~h}\)