# A boat goes \[30km\] upstream and $44km$ downstream in $10$ hours .In $13$ hours it can go $40km$ upstream and $55km$ downstream . Determine the speed of the stream and that of the boat in still water.

$A.$ Speed of boat$ = 8km{\text{ per }}hour$ and speed of stream$ = 3km{\text{ per }}hour$

$B.$ Speed of boat$ = 8km{\text{ per }}hour$ and speed of stream$ = 4km{\text{ per }}hour$

$C.$ Speed of boat$ = 7km{\text{ per }}hour$ and speed of stream$ = 2km{\text{ per }}hour$

$D.$ Data insufficient

Last updated date: 25th Mar 2023

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Hint- This question can be solved by knowing the fact that when a boat goes upstream we subtract the speed of boat and speed of stream and when the boat goes downstream we add the speed of boat and stream.

Now it is given that a boat goes \[30km\] upstream and $44km$ downstream in $10$ hours and in $13$ hours it can go $40km$ upstream and $55km$ downstream.

And we have to find the speed of the stream and speed of the boat in still water.

Let the speed of the boat in still water be $ukm{\text{ per }}hour$ and the speed of stream be $vkm{\text{ per }}hour$.

Now,

Speed of upstream=$\left( {u - v} \right)km{\text{ per }}hour$

Speed of downstream$ = \left( {u + v} \right)km{\text{ per }}hour$

Let $\dfrac{1}{{\left( {u - v} \right)}} = x$ and $\dfrac{1}{{\left( {u + v} \right)}} = y$

Now it is given that a boat goes \[30km\] upstream and $44km$ downstream in $10$ hours.

Or $

\dfrac{{30}}{{u - v}} + \dfrac{{44}}{{u + v}} = 10 \\

\Rightarrow 30x + 44y = 10 - - - - - - - \left( i \right) \\

$

Now multiply $\left( i \right)$ by $4$ we get,

$120x + 176y = 40 - - - - - - \left( {ii} \right)$

Also,

In $13$ hours it can go $40km$ upstream and $55km$ downstream.

$ \Rightarrow \dfrac{{40}}{{u - y}} + \dfrac{{55}}{{u + v}} = 13$

Or $40x + 55y = 13 - - - - - - \left( {iii} \right)$

Now multiply the above equation by 3 we get,

$120x + 165y = 39 - - - - - \left( {iv} \right)$

Subtract $\left( {iv} \right)$ from $\left( {ii} \right)$ we get,

$11y = 1$

Or $y = \dfrac{1}{{11}}$

Now putting the value of $y$ in $\left( i \right)$

$30x + 44 \times \dfrac{1}{{11}} = 10$

Or $30x + 4 = 10$

$

\Rightarrow 30x = 6 \\

\therefore x = \dfrac{1}{5} \\

$

Now we know that $\dfrac{1}{{\left( {u - v} \right)}} = x$ and $\dfrac{1}{{\left( {u + v} \right)}} = y$

Or $u - v = 5$ and $u + v = 11$

Adding the above equation we get the value of u i.e.

$2u = 16$

Or $u = 8km{\text{ per }}hour$

Now put the value of $u$ in any of the above equation,

$8 - 5 = v$

Or $v = 3km{\text{ per }}hour$

Therefore,

Speed of boat in still water$ = 8km{\text{ per }}hour$

Speed of stream$ = 3km{\text{ per }}hour$

Hence, the correct option is $\left( A \right)$ .

Note- Whenever we face such types of questions the key concept is that we should write what is given to us and then convert the statements into equations and then solve them by doing basic mathematics. Like we did in this question.

Now it is given that a boat goes \[30km\] upstream and $44km$ downstream in $10$ hours and in $13$ hours it can go $40km$ upstream and $55km$ downstream.

And we have to find the speed of the stream and speed of the boat in still water.

Let the speed of the boat in still water be $ukm{\text{ per }}hour$ and the speed of stream be $vkm{\text{ per }}hour$.

Now,

Speed of upstream=$\left( {u - v} \right)km{\text{ per }}hour$

Speed of downstream$ = \left( {u + v} \right)km{\text{ per }}hour$

Let $\dfrac{1}{{\left( {u - v} \right)}} = x$ and $\dfrac{1}{{\left( {u + v} \right)}} = y$

Now it is given that a boat goes \[30km\] upstream and $44km$ downstream in $10$ hours.

Or $

\dfrac{{30}}{{u - v}} + \dfrac{{44}}{{u + v}} = 10 \\

\Rightarrow 30x + 44y = 10 - - - - - - - \left( i \right) \\

$

Now multiply $\left( i \right)$ by $4$ we get,

$120x + 176y = 40 - - - - - - \left( {ii} \right)$

Also,

In $13$ hours it can go $40km$ upstream and $55km$ downstream.

$ \Rightarrow \dfrac{{40}}{{u - y}} + \dfrac{{55}}{{u + v}} = 13$

Or $40x + 55y = 13 - - - - - - \left( {iii} \right)$

Now multiply the above equation by 3 we get,

$120x + 165y = 39 - - - - - \left( {iv} \right)$

Subtract $\left( {iv} \right)$ from $\left( {ii} \right)$ we get,

$11y = 1$

Or $y = \dfrac{1}{{11}}$

Now putting the value of $y$ in $\left( i \right)$

$30x + 44 \times \dfrac{1}{{11}} = 10$

Or $30x + 4 = 10$

$

\Rightarrow 30x = 6 \\

\therefore x = \dfrac{1}{5} \\

$

Now we know that $\dfrac{1}{{\left( {u - v} \right)}} = x$ and $\dfrac{1}{{\left( {u + v} \right)}} = y$

Or $u - v = 5$ and $u + v = 11$

Adding the above equation we get the value of u i.e.

$2u = 16$

Or $u = 8km{\text{ per }}hour$

Now put the value of $u$ in any of the above equation,

$8 - 5 = v$

Or $v = 3km{\text{ per }}hour$

Therefore,

Speed of boat in still water$ = 8km{\text{ per }}hour$

Speed of stream$ = 3km{\text{ per }}hour$

Hence, the correct option is $\left( A \right)$ .

Note- Whenever we face such types of questions the key concept is that we should write what is given to us and then convert the statements into equations and then solve them by doing basic mathematics. Like we did in this question.

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