Given:

Downstream distance : Upstream distance = 3 : 1

Speed of stream = 4 km/hr

Downstream time = Upstream time

Concept Used:

Componendo-dividendo Concept

If (a + b)/(a - b) = c/d, then

a/b = (c + d)/(c - d)

Formula Used:

Speed = Distance/Time

D = B + S

U = B - S

where, D → Downstream speed, U → Upstream speed, B → Speed of boat in still water, S → Speed of stream.

Calculations:

Let the time taken for downstream and upstream be t hours each.

Let the distance for downstream and upstream be d₁ and d₂ respectively.

Speed = Distance/Time

D = d₁/t

⇒ (B + S) =  d1/t

⇒ d1 = (B + 4) × t      ----(1)

Similarly, U = d2/t

⇒ (B - S) =  d2/t

⇒ d2 = (B - 4) × t      ----(2)

d₁ : d₂ = 3 : 1

From (1) and (2), we get

[(B + 4) × t]/[(B - 4) × t] = 3/1

⇒ (B + 4)/(B - 4) = 3/1

Using componendo-dividendo, we get

(B + 4) + (B - 4)/(B + 4) - (B - 4) = 3 + 1/3 - 1

⇒ B/4 = 2 / 1

⇒ B = 8 km/hr

∴ The speed of boy in still water is 8 km/hr.

Short Trick/Topper's Approach:

Let the distance for downstream and upstream be d₁ and d2 respectively.

Speed = Distance/Time

For Time = constant,

Speed ∝ Distance

d1 : d2 = 3 : 1

⇒ D : U = 3 : 1 

⇒ (B + S) : (B - S) = 3 : 1

⇒ (B + 4)/(B - 4) = 3/1

Using componendo-dividendo, we get

(B + 4) + (B - 4)/(B + 4) - (B - 4) = 3 + 1/3 - 1

⇒ B/4 = 2 / 1

⇒ B = 8 km/hr

∴ The speed of boy in still water is 8 km/hr.