$A$ 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.

(N/A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h.$
Then the speed of the boat downstream $= (x+y)\, km/h$ and the speed of the boat upstream $= (x-y)\, km/h.$
Using the formula,$\text{time} = \frac{\text{distance}}{\text{speed}}.$
In the first case,the total time taken is $10\, hours$ for $30\, km$ upstream and $44\, km$ downstream:
$\frac{30}{x-y} + \frac{44}{x+y} = 10 \quad ...(1)$
In the second case,the total time taken is $13\, hours$ for $40\, km$ upstream and $55\, km$ downstream:
$\frac{40}{x-y} + \frac{55}{x+y} = 13 \quad ...(2)$
Let $\frac{1}{x-y} = u$ and $\frac{1}{x+y} = v \quad ...(3)$
Substituting these into equations $(1)$ and $(2)$:
$30u + 44v = 10 \quad ...(4)$
$40u + 55v = 13 \quad ...(5)$
Solving these using the elimination method:
Multiply $(4)$ by $4$ and $(5)$ by $3$:
$120u + 176v = 40$
$120u + 165v = 39$
Subtracting the equations: $11v = 1 \implies v = \frac{1}{11}.$
Substituting $v = \frac{1}{11}$ in $(4)$:
$30u + 44(\frac{1}{11}) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{1}{5}.$
Now,$\frac{1}{x-y} = \frac{1}{5} \implies x-y = 5$ and $\frac{1}{x+y} = \frac{1}{11} \implies x+y = 11.$
Adding these: $2x = 16 \implies x = 8.$
Subtracting these: $2y = 6 \implies y = 3.$
Thus,the speed of the boat in still water is $8\, km/h$ and the speed of the stream is $3\, km/h.$