Answer
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Hint: First of all, let the speed of stream be x. Now, the speed of the boat travelling upstream will be $15 - x$ and the speed of the boat travelling downstream will be $15 + x$. Now, as the time taken by boat travelling downstream and travelling upstream is 4 hr 30 minutes, form an equation and use the formula $Time = \dfrac{{dis\tan ce}}{{speed}}$.
Complete step-by-step solution:
In this question, we are given that the speed of the boat in still water is 15 km/hr and it travels 30 km downstream and then returns back to the starting point. The total time taken for this journey is 4 hours 30 minutes. Now, we need to find out the speed of the stream.
Now, let the speed of the stream be x.
Now, while travelling downstream, the speed of the boat will get a boost due to the speed of the stream flowing downwards too. Therefore,
$ \Rightarrow $Speed of boat travelling downstream $ = $ Speed of boat in still water + Speed of stream
$ \Rightarrow $ Speed of boat travelling downstream $ = 15 + x$
So, while travelling upstream, the speed of the boat will reduce due to the speed of the stream. Therefore,
$ \Rightarrow $Speed of boat travelling upstream $ = $ Speed of boat in still water – Speed of stream
$ \Rightarrow $ Speed of boat travelling upstream $ = 15 - x$
Now, the boat travels a distance of 30kms downstream, so it will travel a distance of 30kms only while travelling upstream. Therefore,
$ \Rightarrow $Downstream Distance$ = 30km$
$ \Rightarrow $Upstream Distance$ = 30km$
Now, it is given that the time taken by boat to travel 30kms downstream and then again 30kms upstream is 4 hours 30 minutes. Therefore,
$ \Rightarrow $Time taken to travel downstream + Time taken to travel downstream $ = $4 hour 30 minutes
Now, convert 4 hours 30 minutes into hours first.
$ \Rightarrow $4 hours 30 minutes $ = $ 4.5hr
Now, we know that
$
\Rightarrow Speed = \dfrac{\text{distance}}{\text{time}} \\
\Rightarrow Time = \dfrac{\text{distance}}{\text{speed}} $
Therefore, using this formula, we get
$
\Rightarrow \dfrac{\text{Downstream- distance}}{\text{Downstream- speed}} + \dfrac{\text{Upstream - distance}}{\text{Upstream - speed}} = 4.5 \\
\Rightarrow \dfrac{{30}}{{15 + x}} + \dfrac{{30}}{{15 - x}} = 4.5 $
Taking 30 common and taking LCM on the LHS, we get
$
\Rightarrow 30\left( {\dfrac{1}{{15 + x}} + \dfrac{1}{{15 - x}}} \right) = 4.5 \\
\Rightarrow \dfrac{{15 - x + 15 + x}}{{\left( {15 + x} \right)\left( {15 - x} \right)}} = \dfrac{{4.5}}{{30}} \\
\Rightarrow \dfrac{{30}}{{225 - 15x + 15x - {x^2}}} = 0.15 $
Now, cross multiplying, we get
$
\Rightarrow \dfrac{{30}}{{0.15}} = 225 - 15x + 15x - {x^2} \\
\Rightarrow 200 = 225 - {x^2} $
Gathering all the like terms, we get
$
\Rightarrow {x^2} = 225 - 200 \\
\Rightarrow {x^2} = 25 \\
\Rightarrow x = \pm 5 $
But, speed cannot be negative, so the speed of the stream will be \[x = 5km/hr\].
Note: Note that here the most important is deciding the speeds of the boat while travelling upstream and downstream. When the boat travels upstream, the water is going to oppose the boat so the speed is always going to be less than that of still water. And when the boat is travelling downstream, the speed of the boat is always going to be more than the actual speed as it gets help from the stream that moves downward too.
Complete step-by-step solution:
In this question, we are given that the speed of the boat in still water is 15 km/hr and it travels 30 km downstream and then returns back to the starting point. The total time taken for this journey is 4 hours 30 minutes. Now, we need to find out the speed of the stream.
Now, let the speed of the stream be x.
Now, while travelling downstream, the speed of the boat will get a boost due to the speed of the stream flowing downwards too. Therefore,
$ \Rightarrow $Speed of boat travelling downstream $ = $ Speed of boat in still water + Speed of stream
$ \Rightarrow $ Speed of boat travelling downstream $ = 15 + x$
So, while travelling upstream, the speed of the boat will reduce due to the speed of the stream. Therefore,
$ \Rightarrow $Speed of boat travelling upstream $ = $ Speed of boat in still water – Speed of stream
$ \Rightarrow $ Speed of boat travelling upstream $ = 15 - x$
Now, the boat travels a distance of 30kms downstream, so it will travel a distance of 30kms only while travelling upstream. Therefore,
$ \Rightarrow $Downstream Distance$ = 30km$
$ \Rightarrow $Upstream Distance$ = 30km$
Now, it is given that the time taken by boat to travel 30kms downstream and then again 30kms upstream is 4 hours 30 minutes. Therefore,
$ \Rightarrow $Time taken to travel downstream + Time taken to travel downstream $ = $4 hour 30 minutes
Now, convert 4 hours 30 minutes into hours first.
$ \Rightarrow $4 hours 30 minutes $ = $ 4.5hr
Now, we know that
$
\Rightarrow Speed = \dfrac{\text{distance}}{\text{time}} \\
\Rightarrow Time = \dfrac{\text{distance}}{\text{speed}} $
Therefore, using this formula, we get
$
\Rightarrow \dfrac{\text{Downstream- distance}}{\text{Downstream- speed}} + \dfrac{\text{Upstream - distance}}{\text{Upstream - speed}} = 4.5 \\
\Rightarrow \dfrac{{30}}{{15 + x}} + \dfrac{{30}}{{15 - x}} = 4.5 $
Taking 30 common and taking LCM on the LHS, we get
$
\Rightarrow 30\left( {\dfrac{1}{{15 + x}} + \dfrac{1}{{15 - x}}} \right) = 4.5 \\
\Rightarrow \dfrac{{15 - x + 15 + x}}{{\left( {15 + x} \right)\left( {15 - x} \right)}} = \dfrac{{4.5}}{{30}} \\
\Rightarrow \dfrac{{30}}{{225 - 15x + 15x - {x^2}}} = 0.15 $
Now, cross multiplying, we get
$
\Rightarrow \dfrac{{30}}{{0.15}} = 225 - 15x + 15x - {x^2} \\
\Rightarrow 200 = 225 - {x^2} $
Gathering all the like terms, we get
$
\Rightarrow {x^2} = 225 - 200 \\
\Rightarrow {x^2} = 25 \\
\Rightarrow x = \pm 5 $
But, speed cannot be negative, so the speed of the stream will be \[x = 5km/hr\].
Note: Note that here the most important is deciding the speeds of the boat while travelling upstream and downstream. When the boat travels upstream, the water is going to oppose the boat so the speed is always going to be less than that of still water. And when the boat is travelling downstream, the speed of the boat is always going to be more than the actual speed as it gets help from the stream that moves downward too.
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