A student goes to school at the rate of 21km/h and reaches

A student goes to school at the rate of
2 1 km/h and reaches 6 minutes late.
2
If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is

Let the required distance be x km.

x	−	x	=	16
5/2	−	3	=	60

⇒	2x	−	x	=	4
	5		3		15

⇒	6x − 5x	=	4
	15		15

⇒ x = 4 km.

Second Methdod :

Here, S₁ = 2	1	, t₁ = 6
	2

S₂ = 3, t₂ = 10

Distance =	(S₁ ×S₂)(t₁ + t₂)
	S₂ −S₁

= 15 ×	16	km = 4 km
	60

2	1	km/h and reaches 6 minutes late.
	2