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 $3km/hr$.