Question

A motorboat covers a given distance in 6 hours moving downstream on a river. It covers the same distance in 10 hours moving upstream. The time it takes to cover the same distance in still water is

• A. 9 hours
• B. 7.5 hours
• C. 6.5 hours
• D. 8 hours

Answer 1

The correct option is B 7.5 hours

Let $V_{bw}$ and $V_w$ be the velocities of motorboat and water in the river respectively. Let the distance covered by the motorboat in both the case be $d$.
Case 1: While moving downstream $\frac{d}{6}=V_{bw}+V_w$ .... (i)
Case 2: While moving upstream $\frac{d}{10}=V_{bw}-V_w$ ..... (ii)
Dividing (i) by (ii)
$5/3=(V_{bw}+V_w)/(V_{bw}-V_w)$
$\Rightarrow 5V_{bw}-5V_w=3V_{bw}+3V_w$
$\Rightarrow 2V_{bw}=8V_w$
Time taken in still water $=d/V_{bw}$
Putting value of $V_w$ in (i)
$\frac{d}{6}=V_{bw}+\frac{V_{bw}}{4}\frac{d}{6}=\frac{\left(5V_{bw}\right)}{4}$
$\Rightarrow \frac{d}{V_{bw}}=\frac{5}{2}\times 3=7.5$ hours

Alternative :

Let $t_1$ be the time taken by the motorboat to travel some distance in still water.
Let $t_2$ be the time taken by the motorboat to cover same distance when flowing with the river.
Let $t_3$ and $t_4$ be the time taken by the motorboat to cover same distance while moving downstream and upstream respectively.
$⟹t_3=\frac{\left(t_1{t}_2\right)}{(t_1+t_2)}=6$
and
$t_4=\frac{\left(t_1{t}_2\right)}{(t_2-t_1)}=10$
$\Rightarrow \frac{(t_1+t_2)}{(t_2-t_1)}=\frac{10}{6}=\frac{5}{3}\Rightarrow \frac{t_1}{t_2}=\frac{1}{4}\Rightarrow t_1=\frac{t_2}{4}$
Hence, $6=\frac{\left(4t_1^2\right)}{\left(5t_1\right)}\Rightarrow t_1=7.5$ hours