Let’s assume,

The speed of the boat in still water as $Ckm/hr$

And,

The speed of the stream as $Dkm/hr$

We know that,

Speed of the boat in upstream $=(C–D)km/hr$

Speed of the boat in downstream $=(C+D)km/hr$

So,

Time taken to cover 30 km upstream $=30/(C−D)hr$ [∵ time = distance/ speed]

Time taken to cover 44 km downstream $=44/(C+D)hr$ [∵ time = distance/ speed]

It’s given that the total time of journey is 10 hours. So, this can expressed as

$30/(C–D)+44/(C+D)=10$ …….. (i)

Similarly,

Time taken to cover 40 km upstream $=40/(C–D)hr$ [∵ time = distance/ speed]

Time taken to cover 55 km downstream $=55/(C+D)hr$ [∵ time = distance/ speed]

And for this case the total time of the journey is given as 13 hours.

Hence, we can write

$40/(C–D)+55/(C+D)=13$ ……. (ii)

Hence, CD solving (i) and (ii) we get the required solution

Taking, $1/(C–D)=u$ and $1/(C+D)=v$ in equations (i) and (ii) we have

$30u+44v=10$

$40u+55v=10$

Which can be re- written as,

$30u+44v–10=0$ ……. (iii)

$40u+55v–13=0$……… (iv)

Solving these equations CD cross multiplication we get,

$\frac{u}{44x-13-55x-10}=\frac{-v}{30x-13-40x-10}=\frac{1}{30\times 55-40\times 44}$

$u=\frac{-22}{-110}$

$v=\frac{10}{110}$

$u=\frac{2}{10}$

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

Now,

$1/(C–D)=2/10$

$=1\times C\times 10=2\left( C-D \right)$

⇒ $10=2C–2D$

⇒ $C–D=5$ ……. (v)

And,

$1/(C+D)=1/11$

⇒ $C+D=11$ ……. (vi)

Again, solving (v) and (vi)

Adding (v) and (vi), we get

$2C=16$

⇒ $C=8$

Using C in (v), we find D

$8–D=5$

⇒ $D=3$

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