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A 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 the stream and that of the boat in still water.
Given: A 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: To find the speed of the stream and to find the speed of the boat in still water.
Solution:
Let the speed of the stream $=x\ km/hr$
Let the speed of the boat in still water $=y\ km/hr$
Upstream speed $=y−x\ km/hr$
Downstream speed $=y+x\ km/hr$
$time=\frac{speed}{distance}$
The boat goes $30\ km$ upstream and $44\ km$ downstream in $10\ hours$.
Time taken $=\frac{30}{y−x} +\frac{44}{y+x}$
$\Rightarrow 10= \frac{30}{y−x} +\frac{44}{y+x}$ ................. $( 1)$
The boat goes $40\ km$ upstream and $55\ km$ downstream in $13\ hours$.
Time taken $=\frac{40}{y-x}+\frac{55}{y+x}$
$\Rightarrow 13 =\frac{40}{y-x}+\frac{55}{y+x}$ ................. $( 2)$
Let $\frac{1}{y-x}=u$ and $\frac{1}{y+x}=v$
From $( 1)$ and $( 2)$,
$30u+44v=10$ ................... $( 3)$
$40u+55v=13$ .....................$( 4)$
Multiply equation $( 3)$ with $4$ and equation $( 4)$ with $3$,
$120u+176v=40$ ........... $( 5)$
$120u+165v=39$ ........... $( 6)$
subtract equation $( 6)$ from $( 5)$,
$176v−165v=40−39$
11v=1
$v=\frac{1}{11}$
$\Rightarrow \frac{1}{y+x}=\frac{1}{11}$
$y+x=11$ .................... $( 7)$
From equation $( 3)$,
$30u=10−44v$
$30u=10−44\times \frac{1}{11}$
$30u=10−4=6$
$\Rightarrow u=\frac{6}{30}$
$\Rightarrow u=\frac{1}{5}$
$\Rightarrow y−x=5$ ............... $( 8)$
Adding $( 7)$ and $( 8)$, we get,
$2y=16$
$y=8$
From equation $( 7)$,
$x=11−y$
$x=11−8=3$
Hence, speed of the stream $=x=3\ km/hr$ and speed of the boat in still water $=y=8\ km/hr$
Updated on: 10-Oct-2022
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