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

Given:

The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream.

To do:

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

Solution:

Let the speed of the stream be $x\ km/hr$

Let the speed of the boat in still water be $y\ km/hr$

Upstream speed $=y−x\ km/hr$

Downstream speed $=y+x\ km/hr$

We know that,

$Time=\frac{Speed}{Distance}$

The boat goes $30\ km$ upstream and $44\ km$ downstream in $10\ hours$.

Time taken $=\frac{30}{y−x} +\frac{44}{y+x}$

​$\Rightarrow 10= \frac{30}{y−x} +\frac{44}{y+x}$......(i)

The boat goes $40\ km$ upstream and $55\ km$ downstream in $13\ hours$.

Time taken $=\frac{40}{y-x}+\frac{55}{y+x}$

​$\Rightarrow 13 =\frac{40}{y-x}+\frac{55}{y+x}$........(ii)

Let $\frac{1}{y-x}=u$ and $\frac{1}{y+x}=v$

From (i) and (ii),

$30u+44v=10$......(iii)

$40u+55v=13$.......(iv)

Multiply equation (iii)$ by $4$ and equation (iv) by $3$, we get,

$120u+176v=40$......(v)

$120u+165v=39$......(vi)

Subtracting  equation (vi) from equation (v), we get,

$176v−165v=40−39$

$11v=1$

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

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

$y+x=11$.......(vii)

From equation (iii),

$30u=10−44v$

$30u=10−44\times \frac{1}{11}$

$30u=10−4=6$

$\Rightarrow u=\frac{6}{30}$

$\Rightarrow u=\frac{1}{5}$

$\Rightarrow y−x=5$.....(viii)

Adding equations (vii) and (viii), we get,

$2y=16$

$y=8$

From equation (vii),

$x=11−y$

$x=11−8=3$

Hence, the speed of the stream is $3\ km/hr$ and the speed of the boat in still water is $8\ km/hr$ .

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Updated on: 10-Oct-2022

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