A boy is late by 9 minutes if he walks to school at a speed

A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is

Let the required distance be x km.
According to the question,

x	−	x	=	18
4	−	5	=	60

⇒	5x − 4x	−	3
	20		10

⇒ x =	3	× 20 = 6 km
	10

Second Methdod :
Here, S₁ = 4, t₁ = 9
S₂ = 5, t₂ = 9

Distance =	(S₁ ×S₂)(t₁ + t₂)
	S₂ −S₁

=	(4 × 5)(9 + 9)
	5 − 4

= 20 ×	18	= 6 km
	60