A man can row 40 km upstream and 55 km downstream in 13 hours. He can also row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the boat in still water and the speed of the current.

The correct option is B

8 km/hr, 3 km/hr

Let the speed of boat in still water be x km/hr and speed of current be y km/hr.

Speed upstream = (x - y)km/hr and speed downstream = (x + y)km/hr.

$\frac{40}{x-y}+\frac{55}{x+y}=13and\frac{30}{x-y}+\frac{44}{x+y}=10Let\frac{1}{x-y}=uand\frac{1}{x+y}=v$

The equations reduce to
$40u+55v=13and30u+44v=10$
On writing the equations in standard form, we get
$40u+55v-13=0$
$30u+44v-10=0$
By cross multiplication, we get

$\frac{u}{55\times (-10)-44\times (-13)}=\frac{-v}{40\times (-10)-30\times (-13)}=\frac{1}{44\times 40-55\times 30}\Rightarrow \frac{u}{-550+572}=\frac{-v}{-400+390}=\frac{1}{1760-1650}\Rightarrow \frac{u}{22}=\frac{v}{10}=\frac{1}{110}\Rightarrow u=\frac{22}{110}=\frac{1}{5}andv=\frac{10}{110}=\frac{1}{11}$

So, $u=\frac{1}{5}=\frac{1}{x-y}\Rightarrow x-y=5$ . . . (1)

$andv=\frac{1}{11}=\frac{1}{x+y}\Rightarrow x+y=11$ . . . (2)

On adding (1) & (2), we get

$2x=16$ $\Rightarrow$ $x=8$

and putting $x$ = 8 in (1), we get,

$8-y=5$ $\Rightarrow$ $y=3$.

$\therefore$ Speed of the boat in still water is 8km/hr and speed of the current is 3 km/hr.