Walking at 5 km/hr a student reaches his school from his

Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?

Let the required distance be x km.

Then,	x	−	x	=	24
	3		5		60

⇒	5x − 3x	=	2
	15		5

⇒	2x	= 2
	3

⇒ 2x = 2 × 3 ⇒ x = 3 km

Second Methdod :
Here, S₁ = 3, t₁ = 9
S₂ = 5, t₂ = 15

Distance =	(S₁ ×S₂)(t₁ + t₂)
	S₂ −S₁

=	(3 × 5)(9 + 15)
	5 − 3

=	15	×	24
	2		60

= 3 km