
A boat covers 6 km in an hour in still water. It takes thrice as much time in covering the same distance against the current. What is the speed of the current?
$
(a){\text{ 2km/hr}} \\
(b){\text{ 3km/hr}} \\
(c){\text{ 4km/hr}} \\
(d){\text{ 5km/hr}} \\
$
Answer
511.5k+ views
Hint – In this question let the speed of the boat in still water be x km/hr and speed of boat in current be ykm/hr. Derive relations between these variables by considering the concept of upstream and downstream. Solve the questions to get the answer.
Complete step-by-step solution -
Let the speed of the boat in still water be x km/hr.
And the speed of the current be y km/hr.
So the downstream (D.S) speed = speed of boat + speed of current.
And the upstream (U.S) speed = speed of boat – speed of current.
$ \Rightarrow D.S = x + y$ Km/hr....................... (1)
And
$ \Rightarrow U.S = x - y$ Km/hr......................... (2)
Now as we know the relation of speed, distance and time which is
${\text{Speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$
Let us consider that the distance which covered by the boat be (z) km and let the time taken to cover the distance with the current be (t1) hours and against the current be (t2) hours.
$ \Rightarrow D.S = \dfrac{z}{{{t_1}}}$ Km/hr.
And the upstream speed is 4 km in 2 hours.
$ \Rightarrow U.S = \dfrac{z}{{{t_2}}}$ Km/hr.
$ \Rightarrow D.S\left( {{t_1}} \right) = U.P\left( {{t_2}} \right)$..................... (3)
Now it is given that the boat takes thrice as much time in covering the same distance against the current.
$ \Rightarrow {t_2} = 3{t_1}$
So from equation (3) we have,
$ \Rightarrow D.S\left( {{t_1}} \right) = U.P\left( {3{t_1}} \right)$
$ \Rightarrow D.S = 3\left( {U.S} \right)$
Now from equation (1) and (2) we have,
$ \Rightarrow x + y = 3\left( {x - y} \right)$
Now simplify the above equation we have,
$ \Rightarrow 3x - x = y + 3y$
$ \Rightarrow 4y = 2x$
Now divide by 4 we have,
$ \Rightarrow y = \dfrac{{2x}}{4} = \dfrac{x}{2}$ Km/hr.......................... (4)
Now it is given that a boat covers 6 km in an hour in still water.
So the speed of the boat in still water, x = 6 km/hr.
Now from equation (4) we have,
$ \Rightarrow y = \dfrac{x}{2} = \dfrac{6}{2} = 3$ Km/hr.
So this is the required answer.
Hence option (B) is correct.
Note – In this question the trick part was about the understanding of upstream and downstream, upstream is the direction towards the fluid source or this means that we are going in the opposite direction to the flow as the flow will be directed away from the source. Downstream means towards the direction in which fluid is going or away from the source. That’s why the speed of the boat is added with the speed of current in downstream and subtracted in case of upstream.
.
Complete step-by-step solution -
Let the speed of the boat in still water be x km/hr.
And the speed of the current be y km/hr.
So the downstream (D.S) speed = speed of boat + speed of current.
And the upstream (U.S) speed = speed of boat – speed of current.
$ \Rightarrow D.S = x + y$ Km/hr....................... (1)
And
$ \Rightarrow U.S = x - y$ Km/hr......................... (2)
Now as we know the relation of speed, distance and time which is
${\text{Speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}}$
Let us consider that the distance which covered by the boat be (z) km and let the time taken to cover the distance with the current be (t1) hours and against the current be (t2) hours.
$ \Rightarrow D.S = \dfrac{z}{{{t_1}}}$ Km/hr.
And the upstream speed is 4 km in 2 hours.
$ \Rightarrow U.S = \dfrac{z}{{{t_2}}}$ Km/hr.
$ \Rightarrow D.S\left( {{t_1}} \right) = U.P\left( {{t_2}} \right)$..................... (3)
Now it is given that the boat takes thrice as much time in covering the same distance against the current.
$ \Rightarrow {t_2} = 3{t_1}$
So from equation (3) we have,
$ \Rightarrow D.S\left( {{t_1}} \right) = U.P\left( {3{t_1}} \right)$
$ \Rightarrow D.S = 3\left( {U.S} \right)$
Now from equation (1) and (2) we have,
$ \Rightarrow x + y = 3\left( {x - y} \right)$
Now simplify the above equation we have,
$ \Rightarrow 3x - x = y + 3y$
$ \Rightarrow 4y = 2x$
Now divide by 4 we have,
$ \Rightarrow y = \dfrac{{2x}}{4} = \dfrac{x}{2}$ Km/hr.......................... (4)
Now it is given that a boat covers 6 km in an hour in still water.
So the speed of the boat in still water, x = 6 km/hr.
Now from equation (4) we have,
$ \Rightarrow y = \dfrac{x}{2} = \dfrac{6}{2} = 3$ Km/hr.
So this is the required answer.
Hence option (B) is correct.
Note – In this question the trick part was about the understanding of upstream and downstream, upstream is the direction towards the fluid source or this means that we are going in the opposite direction to the flow as the flow will be directed away from the source. Downstream means towards the direction in which fluid is going or away from the source. That’s why the speed of the boat is added with the speed of current in downstream and subtracted in case of upstream.
.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
For Frost what do fire and ice stand for Here are some class 10 english CBSE

What did the military generals do How did their attitude class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

List three states in India where earthquakes are more class 10 physics CBSE

What did being free mean to Mandela as a boy and as class 10 english CBSE

Where did the fight between the two campaigns of Sambhaji class 10 social science CBSE
