A motorboat whose speed is 20 km/h in still water takes 30 minutes more to go 24 km upstream than to cover the same distance downstream. If the speed of the boat in still water is increased by 2 km/h, then how much time will it take to go 39 km downstream and 30 km upstream?

Given:

The speed of the motorboat in still water = 20 km/h

Concept used:

If the speed of a boat in still water is x km/h and the speed of the stream is y km/h, then

Downstream speed = (x + y) km/h

Upstream speed = (x - y) km/h

Time = Distance/Speed

Calculation:

According to the question, the motorboat takes 30 minutes more to go 24 km upstream than to cover the same distance downstream.

Let, the speed of the water = x km/h

So, 24/(20 - x) = 24/(20 + x) + (1/2)  [∵ 30 minutes = 1/2 hour]

⇒ 24/(20 - x) - 24/(20 + x) = (1/2)

⇒ 

⇒ 

⇒ 

⇒ 400 - x² = 96x

⇒ x² + 96x - 400 = 0

⇒ x2 + 100x - 4x - 400 = 0

⇒ x (x + 100) - 4 (x + 100) = 0

⇒ (x + 100) (x - 4) = 0

⇒ x + 100 = 0 ⇒ x = -100 ["-" is neglacted]

⇒ x - 4 = 0 ⇒ x = 4

∴ The speed of the water = 4 km/h

The speed of the motorboat in still water increased 2 km/h = 20 + 2 = 22 km/h

The time for 39 km downstream and 30 km upstream = 39/(22 + 4) + 30/(22 - 4) hours

= (39/26) + (30/18) hours

= 3/2 + 5/3 hours

= 19/6 hours

= (19/6) × 60 minutes

= 190 minutes

= 3 hours 10 minutes

∴ The motorboat will take 3 hours 10 minutes to go 39 km downstream and 30 km upstream

Shortcut TrickValue putting method, 

According to the question, 

30 min = 1/2 hr

x = 20 (Speed in still water)

⇒ 24/(20 - y) - 24/(20 + y) = 1/2

Here the R.H.S is 1/2, so the value of 20 - y must be more than 12

Hence take y = 4 (so that right bracket will become 1 as 20 + 4 = 24) and (left bracket will be more than half)

⇒ 24/(20 - 4) - 24(20 + 4) = 3/2 - 1 = 1/2

Hence the value of Y = 4

Now according to the question, 

⇒ 39/(22 + 4) + 30/(22 - 4) = 39/26 + 30/18

⇒ 19/6 = 3(1/6) = 3 hours and 10 min

∴ The motorboat will take 3 hours 10 minutes to go 39 km downstream and 30 km upstream