Question : A man rows from J to K (upstream) and back from K to J (downstream) in a total time of 15 hours. The distance between J and K is 300 km. The time taken by the man to row 9 km downstream is identical to the time taken by him to row 3 km upstream. What is the approximate speed of the boat in still water?
Option 1: 51.33 km/hr
Option 2: 47.67 km/hr
Option 3: 53.33 km/hr
Option 4: 43.67 km/hr
Correct Answer: 53.33 km/hr
Solution :
Let the speed of the boat as B and the speed of the stream as S.
Upstream Speed = B – S
Downstream Speed = B + S
⇒ A man rows from J to K (upstream) and back from K to J(downstream) in a total time of 15 hours. The distance between J and K is 300 km.
According to the question,
$\frac{300}{B + S}$ + $\frac{300}{B – S}$ = 15
⇒ $\frac{20}{B + S}$ + $\frac{20}{B – S}$ = 1
Since the time taken by him to row 3 km upstream is equal to the time taken by him to row 9 km downstream.
So, $\frac{9}{B + S}$ = $\frac{3}{B – S}$
⇒ $\frac{B + S}{B - S}$ = $\frac{3}{1}$
From this, we can take the speed of the boat as B = 2$x$ km/hr, and the speed of the stream as S = $x$ km/hr
So, $\frac{20}{ 2x + x} + \frac{20}{2x – x}=1$
⇒ $\frac{20}{ 3x } + \frac{20}{x}=1$
⇒ $x = \frac{80}{3}$
So, the speed of boat in still water, B = 2$x$ km/hr = 2 × $\frac{80}{3}$ = 53.33 km/hr
Hence, the correct answer is 53.33 km/hr.
Related Questions
Know More about
Staff Selection Commission Multi Tasking ...
Application | Cutoff | Selection Process | Preparation Tips | Eligibility | Exam Pattern | Admit Card
Get Updates Brochure
Your Staff Selection Commission Multi Tasking Staff Exam brochure has been successfully mailed to your registered email id “”.
