Question

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

Answer 1

Let the speed of the boat in still water be $xkm/hr$ and the speed of the stream be $ykm/hr$

Speed upstream $=(x-y)km/hr$

Speed downstream $=(x+y)km/hr$

Now,

Time taken to cover $30km$ upstream $=\frac{30}{x-y}hrs$

Time taken to cover $44km$ downstream $=\frac{44}{x+y}hrs$

But total time of journey is $10hours$

$\frac{30}{x-y}+\frac{44}{x+y}=10\dots \dots \left(i\right)$

Time taken to cover $40km$ upstream $=\frac{40}{x-y}hrs$

Time taken to cover $55km$ down stream $=\frac{55}{x+y}hrs$

In this case total time of journey is given to be $13hours$

Therefore,$\frac{40}{x-y}+\frac{55}{x+y}=13\dots \dots \left(ii\right)$

Putting $\frac{1}{x-y}=u$ and $\frac{1}{x+y}=v$ in equation (i) and (ii), we get

$30u+44v=10$

$40u+55v=13$

$30u+44v-10=0\dots \dots \left(iii\right)$

$40u+55v-13=0\dots \dots \left(iv\right)$

Solving these equations by cross multiplication we get

$\frac{u}{44\times -13-55\times -10}=\frac{-v}{30\times -13-40\times -10}=\frac{1}{30\times 55-40\times 44}$

$\frac{u}{-572+550}=\frac{-v}{-390+400}=\frac{1}{1650-1760}$

$\frac{u}{-22}=\frac{-v}{10}=\frac{1}{-110}$

$u=\frac{-22}{-110}=\frac{2}{10}$

and, $v=\frac{-10}{-110}=\frac{1}{11}$

Now,

$u=\frac{2}{10}$

$\frac{1}{x-y}=\frac{2}{10}$

$1\times 10=2(x-y)$

$x-y=5\dots \dots \left(v\right)$

And, $v=\frac{1}{11}$

$\frac{1}{x+y}=\frac{1}{11}$

$x+y=11\dots \dots \left(vi\right)$

By solving equation $\left(v\right)$ and $\left(vi\right),$ we get

$(x-y)+(x+y)=5+11$

$2x=16$

$x=8$

Substituting $x=8$ in equation $\left(vi\right)$, we get

$x+y=11$

$8+y=11$

$y=11-8=3$

Hence, the speed of the boat in still water is $8km/hr$

The speed of the stream is $3km/hr.$