Speed \(=\) distance \(/\) time
Let the speed of boat be ' \(a\) ' and speed of stream be ' \(b\) '
Relative speed of boat going upstream \(=a-b\)
Relative speed of boat going downstream \(=a+b\)
Given, boat goes \(30 \mathrm{~km}\) upstream and \(44 \mathrm{~km}\) downstream in 10 hours.
In 13 hours, it can go \(40 \mathrm{~km}\) upstream and \(55 \mathrm{~km}\) down stream and we knov
$$
\begin{aligned}
&\text { time }=\frac{\text { speed }}{\text { distance }} \\
&\Rightarrow \frac{30}{a-b}+\frac{44}{a+b}=10 \cdots(1)
\end{aligned}
$$
Also,
$$
\frac{40}{a-b}+\frac{55}{a+b}=13 \cdots(2)
$$
Now take,
$$
\frac{1}{a-b}=u, \frac{1}{a+b}=v
$$
Eq 1 and 2 becomes
$$
\Rightarrow 30 u+44 v=10
$$
Take 2 common out of above equation
$$
\Rightarrow 15 u+22 v-5=0 \ldots . \text { (3) }
$$
and
$$
40 u+55 v-13=0 \ldots .(4)
$$
Solve the equations by cross multiplication method
$$
\begin{aligned}
& \frac{u}{22 \times(-13)-55 \times(-5)} \\
=& \frac{-v}{15 \times(-13)-40 \times(-5)} \\
=& \frac{1}{15 \times(55)-40 \times(22)} \\
\frac{u}{-286+275}=& \frac{-v}{-195+200} \\
=& \frac{1}{825-880}
\end{aligned}
$$
$$
\begin{aligned}
&\frac{u}{-11}=\frac{-v}{5}=\frac{1}{-55} \\
&u=\frac{-11}{-55}, v=\frac{-5}{-55} \\
&u=\frac{1}{5}, v=\frac{1}{11} \\
&\Rightarrow \mathrm{a}-\mathrm{b}=5 \ldots .(4) \\
&\mathrm{a}+\mathrm{b}=11
\end{aligned}
$$
Add eq 4 and 5 to get
$$
\begin{aligned}
&a-b+a+b=5+11 \\
&\Rightarrow 2 a=16 \\
&\Rightarrow a=8
\end{aligned}
$$
Substitute value of a in eq 4 we get,
$$
\begin{aligned}
&8-b=5 \\
&\Rightarrow b=8-5 \\
&\Rightarrow b=3
\end{aligned}
$$
Speed of boat \(=8 \mathrm{~km} / \mathrm{hr}\) and speed of stream \(=3 \mathrm{~km} / \mathrm{hr}\)
