Let's assume,
The speed of the boat In still water as \(\mathrm{x} \mathrm{km} / \mathrm{hr}\)
And,
The speed of the stream as \(\mathrm{y} \mathrm{km} / \mathrm{hr}\)
We know that,
Speed of the boat In upstream \(=(x-y) \mathrm{km} / \mathrm{hr}\)
Speed of the boat In downstream \(=(x+y) \mathrm{km} / \mathrm{hr}\)
So, time taken to cover \(28 \mathrm{~km}\) downstream \(=28 /(x+y)\) hr [ : time = distance/ speed]
Time taken to cover \(24 \mathrm{~km}\) upstream \(=24 /(x-y)\) hr [ : tIme = distance/ speed]
It's given that the total time of journey is 6 hours.
So, this can expressed as \(24 /(x-y)+28 /(x+y)=6 \ldots \ldots\) (i)
Similarly,
Time taken to cover \(30 \mathrm{~km}\) upstream \(=30 /(x-y)[-\) time \(=\) distance/ speed \(]\)
Time taken to cover \(21 \mathrm{~km}\) downstream \(=21 /(x+y)\) [: time = distance/ speed]
And for this case the total time of the journey is given as \(6.5\) i.e \(13 / 2\) hours.
Hence, we can write
\(30 /(x-y)+21 /(x+y)=13 / 2 \ldots\). (ii)Hence, by solving (I) and (II) we get the required solutlon
Taking, \(1 /(x-y)=u\) and \(1 /(x+y)=v\) in equations (i) and (ii) we have (after rearranging)
\(24 u+28 v-6=0 \ldots\) (iii)
\(30 u+21 v-13 / 2=0 \ldots \ldots\) (iv)
Solving these equatlons by cross multiplicatlon we get,
$$
\begin{aligned}
&\frac{v}{28 z-6.5 \cdot 21 z-6}=\frac{v}{24 x-6.50 z 6}=\frac{1}{24 \times 2130 \times 28} \\
&u=1 / 6 \text { and } v=1 / 14
\end{aligned}
$$
Now,
$$
\begin{aligned}
&u=1 /(x-y)=1 / 6 \\
&x-y=6 \ldots(v) \\
&v=1 /(x+y)=1 / 14 \\
&x+y=14 \ldots \ldots(v i)
\end{aligned}
$$
On Solving (v) and (vI)
Adding (v) and (vI), we get
$$
\begin{aligned}
&2 x=20 \\
&\Rightarrow x=10
\end{aligned}
$$
Using \(x=10\) in (v), we find \(y\)
$$
\begin{aligned}
&10+y=14 \\
&\Rightarrow y=4
\end{aligned}
$$
Therefore,
speed of the stream \(=4 \mathrm{~km} / \mathrm{hr}\).
Speed of boat \(=10 \mathrm{~km} / \mathrm{hr}\).
