A boy reduces his speed to \(\frac{7}{11}\) and reaches his destination in 22 hours. How much time he will save if he walk at his initial speed?

8 hrs

S₁ = Actual Speed

S₂ = New Speed

T₁ = Actual Time

T₂ = New Time

*Time ∝ Speed when distance is same*

ATQ,

S₁ : S₂ = 11 : 7

T₁ : T₂ =  7 : 11  ⇒ 4R (Saved time at normal speed)

11R = 22hrs (given)

1R = 2hrs.

Time save at normal speed = 4R = 4× 2 = 8 hrs.