# A motor boat whose speed is 18 km/hr in still water 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream

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Hint: First of all, let the speed of the stream be $'u'\ km/hr$. Then we get the downstream speed of the boat $=\left( 18+u \right)km/hr$ and upstream speed $=\left( 18-u \right)\ km/hr$. Then use,

(time taken by boat to go 24 km upstream) – (time taken by boat to go 24 km downstream) = 1

Here, find time by using $\text{time}\ =\dfrac{\text{Distance}}{\text{Speed}}$

Complete step-by-step answer:

Here, we are given a motor boat whose speed is 18 km/hr in still water. It takes 1 hour more to go 24 km upstream than to return downstream to the same spot. We have to find the speed of the stream.

Let us consider the speed of the stream to be $'u'\ km/hr$.

We know that whenever a boat goes downstream, then the direction of stream and direction of boat is in the same direction. Hence stream supports to boat while going down stream so, we get,

downstream speed of boat = speed of boat in still water + speed of stream........................(i)

We also know that whenever a boat goes upstream then the direction of stream and direction of boat is in the opposite direction.

Hence, the stream opposes the boat while going upstream. So we get,

Upstream speed of boat = speed of boat in still water – speed of stream................(ii)

As we are given that speed of motor is still water as 18 km/hr and we have assumed speed of stream as$'u'\ km/hr$, by putting these in equation (i) and equation (ii), we get,

Downstream speed of boat = (18 + u) km/hr

and upstream speed of boat = (18 – u) km/hr

Now, let us consider the line taken by boat to go 24 km downstream as td.

Since, we know that, $\text{time =}\dfrac{\text{distance}}{\text{speed}}$, therefore by

Putting the value of time = td, distance = 24 km and downstream speed \[=\left( 18+u \right)km/hr\] , we get

$td=\dfrac{24km}{\left( 18+u \right)km/hr}.....................(iii)$

Now, let us consider the time taken by boat to go 24 km upstream as tu.

The value of time = tu, distance = 24 km and upstream speed \[=\left( 18-u \right)km/hr\], we get,

$tu=\dfrac{24km}{\left( 18-u \right)km/hr}................................(iv)$

Now we are give that time taken by boat to go 24 km upstream is greater than time taken by boat to go 24 km downstream by 1hour, therefore we get,

tu – td = 1 hour

By putting the value of td and tu from equation (iii) and equation (iv) respectively in above equation, we get,

$\dfrac{24}{\left( 18-u \right)}-\dfrac{24}{\left( 18+u \right)}=1$

By taking 24 common and simplifying above equation, we get,

$\begin{align}

& 24\left[ \dfrac{\left( 18+u \right)-\left( 18-u \right)}{\left( 18+u \right)\left( 18-u \right)} \right]=1 \\

& \Rightarrow 24\left[ \dfrac{u+u}{\left( 18+u \right)\left( 18-u \right)} \right]=1 \\

\end{align}$

By cross multiplying above equation, we get,

$24\left( 2u \right)=\left( 18+u \right)\left( 18-u \right)$

We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. By applying this in above equation, we get,

$\begin{align}

& 48u={{\left( 18 \right)}^{2}}-{{u}^{2}} \\

& or\ {{u}^{2}}+48u-324=0 \\

\end{align}$

Here we can split $48u=54u-6u$

Therefore we get,

\[{{u}^{2}}+54u-6u-324=0\]

We can also write it as,

\[u\left( u+54 \right)-6\left( u+54 \right)=0\]

By taking $\left( u+54 \right)$common, we get,

$\left( u+54 \right)\left( u-6 \right)=0$

Therefore we get $u=-54$and $u=6$.

As we have assumed that $'u'$ is the speed of stream it can’t be negative. Hence we get $u=6km/hr$.

Therefore, we get a speed of the stream $6km/hr$.

Note: If we have speed of boat in still water as $'v'km/hr$ and speed of stream as $'u'km/hr$, some students often make this mistake of taking downstream speed as $\left( u+v \right)km/hr$ and upstream speed as $\left( u-v \right)km/hr$. While the downstream can speed is correct but upstream speed is wrong. The correct upstream speed of the boat is $\left( v-u \right)km/hr.$ so this mistake must be avoided. Students can remember it like, while going upstream, the stream opposes the flow of the boat so the speed of the stream must get subtracted from the speed of the boat and not the other way around.

(time taken by boat to go 24 km upstream) – (time taken by boat to go 24 km downstream) = 1

Here, find time by using $\text{time}\ =\dfrac{\text{Distance}}{\text{Speed}}$

Complete step-by-step answer:

Here, we are given a motor boat whose speed is 18 km/hr in still water. It takes 1 hour more to go 24 km upstream than to return downstream to the same spot. We have to find the speed of the stream.

Let us consider the speed of the stream to be $'u'\ km/hr$.

We know that whenever a boat goes downstream, then the direction of stream and direction of boat is in the same direction. Hence stream supports to boat while going down stream so, we get,

downstream speed of boat = speed of boat in still water + speed of stream........................(i)

We also know that whenever a boat goes upstream then the direction of stream and direction of boat is in the opposite direction.

Hence, the stream opposes the boat while going upstream. So we get,

Upstream speed of boat = speed of boat in still water – speed of stream................(ii)

As we are given that speed of motor is still water as 18 km/hr and we have assumed speed of stream as$'u'\ km/hr$, by putting these in equation (i) and equation (ii), we get,

Downstream speed of boat = (18 + u) km/hr

and upstream speed of boat = (18 – u) km/hr

Now, let us consider the line taken by boat to go 24 km downstream as td.

Since, we know that, $\text{time =}\dfrac{\text{distance}}{\text{speed}}$, therefore by

Putting the value of time = td, distance = 24 km and downstream speed \[=\left( 18+u \right)km/hr\] , we get

$td=\dfrac{24km}{\left( 18+u \right)km/hr}.....................(iii)$

Now, let us consider the time taken by boat to go 24 km upstream as tu.

The value of time = tu, distance = 24 km and upstream speed \[=\left( 18-u \right)km/hr\], we get,

$tu=\dfrac{24km}{\left( 18-u \right)km/hr}................................(iv)$

Now we are give that time taken by boat to go 24 km upstream is greater than time taken by boat to go 24 km downstream by 1hour, therefore we get,

tu – td = 1 hour

By putting the value of td and tu from equation (iii) and equation (iv) respectively in above equation, we get,

$\dfrac{24}{\left( 18-u \right)}-\dfrac{24}{\left( 18+u \right)}=1$

By taking 24 common and simplifying above equation, we get,

$\begin{align}

& 24\left[ \dfrac{\left( 18+u \right)-\left( 18-u \right)}{\left( 18+u \right)\left( 18-u \right)} \right]=1 \\

& \Rightarrow 24\left[ \dfrac{u+u}{\left( 18+u \right)\left( 18-u \right)} \right]=1 \\

\end{align}$

By cross multiplying above equation, we get,

$24\left( 2u \right)=\left( 18+u \right)\left( 18-u \right)$

We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. By applying this in above equation, we get,

$\begin{align}

& 48u={{\left( 18 \right)}^{2}}-{{u}^{2}} \\

& or\ {{u}^{2}}+48u-324=0 \\

\end{align}$

Here we can split $48u=54u-6u$

Therefore we get,

\[{{u}^{2}}+54u-6u-324=0\]

We can also write it as,

\[u\left( u+54 \right)-6\left( u+54 \right)=0\]

By taking $\left( u+54 \right)$common, we get,

$\left( u+54 \right)\left( u-6 \right)=0$

Therefore we get $u=-54$and $u=6$.

As we have assumed that $'u'$ is the speed of stream it can’t be negative. Hence we get $u=6km/hr$.

Therefore, we get a speed of the stream $6km/hr$.

Note: If we have speed of boat in still water as $'v'km/hr$ and speed of stream as $'u'km/hr$, some students often make this mistake of taking downstream speed as $\left( u+v \right)km/hr$ and upstream speed as $\left( u-v \right)km/hr$. While the downstream can speed is correct but upstream speed is wrong. The correct upstream speed of the boat is $\left( v-u \right)km/hr.$ so this mistake must be avoided. Students can remember it like, while going upstream, the stream opposes the flow of the boat so the speed of the stream must get subtracted from the speed of the boat and not the other way around.

Last updated date: 21st Sep 2023

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