The distance between A and B is 30 km. A boat makes a return journey from point A to point B and back in 10 hours 30 minutes. One way it travels with the stream and on the return it travels against the stream. If the speed of the stream increases by 2 km/hr, the return journey takes 17 hours 30 minutes. What is the speed of the boat in still water?

Question: The distance between A and B is 30 km. A boat makes a return journey from point A to point B and back in 10 hours 30 minutes. One way it travels with the stream and on the return it travels against the stream. If the speed of the stream increases by 2 km/hr, the return journey takes 17 hours 30 minutes. What is the speed of the boat in still water?

Solution:
Let,
x km/hr be the speed of boat upstream
y km/hr be the speed of boat downstream

∴ 30/x + 30/y = 10 hours 30 minutes = 10.5 hour = 105/10 hour = 21/2
⇒ 30/x + 30/y = 21/2
⇒ 60/x + 60/y = 21
⇒ 20/x + 20/y = 7
⇒ 20/x = 7 - 20/y
⇒ 20/x = (7y - 20)/y
⇒ x/20 = y/(7y - 20)
∴ x = 20y/(7y - 20) ............(1)


If the speed of the stream increases by 2 km/hr
30/(x - 2) + 30/(y + 2) = 17 hours 30 minutes = 17.5 hour = 175/10 = 35/2
⇒ 30/(x - 2) + 30/(y + 2) = 35/2
⇒ 60/(x - 2) + 60/(y + 2) = 35
⇒ 12/(x - 2) + 12/(y + 2) = 7
⇒ 12/(x - 2) = 7 - 12/(y + 2)
⇒ 12/(x - 2) = (7y + 14 -12)/(y + 2)
⇒ 12/(x - 2) = (7y + 2)/(y + 2)
⇒ (x - 2)/12 = (y + 2)/(7y + 2)
⇒ x - 2 = (12y + 24)/(7y + 2)
⇒ x = (12y + 24)/(7y + 2) + 2
⇒ x = (12y + 24 + 14y + 4)/(7y + 2)
∴ x = (26y + 28)/(7y + 2) ...............(2)

From (1) and (2) we get,
20y/(7y - 20) = (26y + 28)/(7y + 2)
⇒ 20y(7y + 2) = (26y + 28)(7y - 20)
⇒ 140y² + 40y = 182y² - 324y - 560
⇒ 42y² - 364y - 560 = 0
⇒ 3y² - 26y - 40 = 0
⇒ 3y² - 30y + 4y - 40 = 0
⇒ 3y(y - 10) + 4(y - 10) = 0
⇒ (y - 10)(3y + 4) = 0
∴ y = 10 or y = - 4/3 [which is not acceptable]
∴ y =10

From (1) we get
x = 20y/(7y - 20)
= 200/(70 - 20)
= 4

∴ The speed of the boat in still water = (10 + 4)/2 km/hr
= 7 km/hr