Question

a boat goes 24 km upstream and 28 km downstream in 6 hours it goes 30 km upstream and 21 km downstream in 6 hours and 30 minutes.find the boat in stil water​

Answer 1

AnSwer:

Let speed of boat in still water be x km/hr.

Let present rate be y km/hr.

∴ Speed of the boat upstream =(x-y) km/hr.

★ According to the question,

$\sf \longmapsto \: \: \dfrac{24}{x - y} + \dfrac{28}{x + y} = 6 \: \: - - - - (1) \\ \\ \\ \sf \longmapsto \: \: \dfrac{30}{x - y} + \dfrac{21}{x + y} = \dfrac{13}{2} \: \: - - - - (2)$

Now,

From eqn (1) and (2) respectively,

$\sf \longmapsto \: \: \dfrac{1}{x - y} = \dfrac{1}{6} \: and \: \dfrac{1}{x + y} = \dfrac{1}{14} \\ \\ \sf \longmapsto \: \: x - y = 6 \: - - (3) \\ \\ \sf \longmapsto \: \: x+ y = 14 - - (4)$

Now by substitution method,

Considering eqn (3),

$\sf \longmapsto \: \:x = 6 + y - - (5)$

Substitute the value of x in eqn (4),

$\sf \longmapsto \: \:x + y = 14 \\ \\ \sf \longmapsto \: \:(6 + y) + y = 14 \\ \\ \sf \longmapsto \: \:6 + 2y = 14 \\ \\ \sf \longmapsto \: \:2y = 14 - 6 \\ \\ \sf \longmapsto \: \:y = \dfrac{8}{2} \\ \\ \longmapsto { \underline{ \boxed{ \purple{ \sf{\: \:y = 4}}}}} \: \bigstar$

Similarly, substitute the value of y = 4 in eqn (5),

$\sf \longmapsto \: \:x = 6 + y \\ \\ \sf \longmapsto \: \:x = 6 + 4 \\ \\ \sf \longmapsto { \underline{ \boxed{ \red{ \sf{\: \:x = 10}}}}} \: \bigstar$

Present rate is 4 km/hr.

So, the speed of boat in still water is 10 km/hr.