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
$$
\begin{gathered}
\left(\mathrm{v}_{\mathrm{b}}+\mathrm{v}_{\mathrm{w}}\right) \times 6=\left(\mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{w}}\right) \times 10 \\
\mathrm{v}_{\mathrm{w}}=\frac{\mathrm{v}_{\mathrm{b}}}{4} \\
\therefore \quad \mathrm{x}=\left(\mathrm{v}_{\mathrm{b}}+\mathrm{v}_{\mathrm{w}}\right) \times 6=7.5 \mathrm{v}_{\mathrm{b}}
\end{gathered}
$$
Time taken by the motorboat to cover the same distance in still water is
$$
t=\frac{x}{v_{b}}=\frac{7.5 v_{b}}{v_{b}}=7.5 \mathrm{~h}
$$