Question

A boat takes 2 hours to go 8 km and comes back in a still water lake. The time taken for going 8 km upstream and coming back with water velocity of $4\dfrac{{km}}{{hr.}}$ is:
A) 140 min.
B) 150 min.
C) 160 min.
D) 170 min.

Answer 1

Hint: The speed of water flowing can be added to the speed of the boat while going downstream and the speed of the flowing water can be subtracted from the speed of the boat as it is going upstream and by applying a simple formula of speed the time taken can be calculated.

Formula used:
The formula of the speed is given by,
${\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$.

Complete step by step answer:
It is given in the problem that a boat takes 2 hours to go 8 km and comes back in still water lake and we need to find the time taken for going 8 km upstream and coming back with water velocity of $4\dfrac{{km}}{{hr.}}$.
First of all let us calculate the speed of the boat in still water.
$ \Rightarrow {\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
The distance here is 16 km and the time taken is 2 hrs. then the speed of the boat will be,
$ \Rightarrow {\text{speed}} = \dfrac{{16}}{2}$
$ \Rightarrow {\text{speed}} = 8\dfrac{{km}}{{hr}}$
Now when the boat is going upstream then the speed of the boat will be,
$ \Rightarrow {v_p} = \left( {8 - 4} \right)\dfrac{{km}}{{hr.}}$
$ \Rightarrow {v_p} = 4\dfrac{{km}}{{hr.}}$
Therefore the time taken will be equal to,
$ \Rightarrow {\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
$ \Rightarrow {\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}}$
$ \Rightarrow {\text{time}} = \dfrac{8}{4}hr.$
$ \Rightarrow {{\text{t}}_p} = 2hr.$………eq. (1)
Now the speed for the boat in downstream is equal to,
${v_d} = \left( {8 + 4} \right)\dfrac{{km}}{{hr.}}$
${v_d} = 12\dfrac{{km}}{{hr.}}$
The time taken in downstream is equal to,
$ \Rightarrow {\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}}$
$ \Rightarrow {{\text{t}}_d} = \dfrac{{\text{8}}}{{12}}hr.$
$ \Rightarrow {{\text{t}}_d} = 0 \cdot 67hr.$………eq. (2)
The total time taken will be equal to,
$ \Rightarrow {t_t} = {t_p} + {t_d}$
On substituting the corresponding values,
$ \Rightarrow {t_t} = 2 + 0 \cdot 67$
$ \Rightarrow {t_t} = 2 \cdot 67hr.$
On simplification of the above equation,
$ \Rightarrow {t_t} = \left( {2 \cdot 67} \right) \times \left( {60} \right)\min .$
On further simplification, we get
$ \Rightarrow {t_t} = 160.2\min $
$ \Rightarrow {t_t} \approx 160\min .$

The time taken in going upstream and coming back is ${t_t} = 160\min $. The correct option for this problem is option C.

Note:
The time taken by the boat going upstream will be more because the flow of the water will resist the motion of the boat whereas if the boat is going down the stream then the speed of the boat will be increased and the time taken will be less as the flow of water will help the motion of the boat.