A student rides on bicycle at 8 km/hour and reaches his

A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?

Let x km. be the required distance.
Difference in time = 2.5 + 5 = 7.5 minutes

=	7.5	hrs. =	1	hrs.
	60		8

Now,	x	−	x	=	1
	8		10		8

⇒	5x − 4x	=	1
	40		8

⇒ x =	40	= 5 km.
	8

Second Methdod :
Here, S₁ = 8, t₁ = 2.5
S₂ = 10, t₂ = 5

Distance =	(S₁ ×S₂)(t₁ + t₂)
	S₂ −S₁

=	(8 × 10)(2.5 + 5)
	10 − 8

= 40 ×	7.5	= 5 km
	60