A student goes to school at the rate of5km/hr and reaches

A student goes to school at the
rate of 5 km/hr and reaches 6 minutes late.
2
If he travels at the speed of 3 km/hr, he reaches 10 minutes earlier. The distance of the school is

Distance of school = x km
Difference of time

= 16 minutes =	16	hour
	60

∴	x	−	x	=	16
	5/2		3		60

⇒	2x	−	x	=	4
	5		3		15

⇒	6x − 5x	=	4
	15		15

⇒	x	=	4
	15		15

⇒ x =	4	× 15 = 4km
	15

Second Method :

Here, S₁ =	5	,t₁ = 6
	2

S₂ = 3, t₂ = 10

Distance =	(S₁ × S₂)(t₁ + t₂)
	S₁ − S₂

= 15 ×	16	km = 4 km.
	60