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# A motor boat can travel $ 30 \mathrm{~km} $ upstream and $ 28 \mathrm{~km} $ downstream in 7 hours. It can travel $ 21 \mathrm{~km} $ upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Given:

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours.

To do:

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

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 $28\ km$ downstream in $7\ hours$.

Time taken $=\frac{30}{y−x} +\frac{28}{y+x}$

$\Rightarrow 7= \frac{30}{y−x} +\frac{28}{y+x}$......(i)

The boat goes $21\ km$ upstream and return in 5 hours.

Time taken $=\frac{21}{y-x}+\frac{21}{y+x}$

$\Rightarrow 5 =\frac{21}{y-x}+\frac{21}{y+x}$........(ii)

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

From (i) and (ii),

$30u+28v=7$......(iii)

$21u+21v=5$.......(iv)

Multiply equation (iii) by $7$ and equation (iv) by $10$, we get,

$7(30u+28v)=7(7)$

$210u+196v=49$......(v)

$10(21u+21v)=10(5)$

$210u+210v=50$......(vi)

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

$210v−196v=50−49$

$14v=1$

$v=\frac{1}{14}$

$\Rightarrow \frac{1}{y+x}=\frac{1}{14}$

$y+x=14$.......(vii)

From equation (iii),

$30u=7−28v$

$30u=7−28\times \frac{1}{14}$

$30u=7−2=5$

$\Rightarrow u=\frac{5}{30}$

$\Rightarrow u=\frac{1}{6}$

$\Rightarrow y−x=6$.....(viii)

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

$2y=20$

$y=10$

From equation (vii),

$x=14−y$

$x=14−10=4$

Hence, the speed of the stream is $4\ km/hr$ and the speed of the boat in still water is $10\ km/hr$ .

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