Walking at a speed of 5 km/hr, a man reaches his office

Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is

Let the distance of the office be x km, then

x	−	x	=	8
5	−	6	=	60

⇒	6x − 5x	=	2
	30		15

⇒ x = 2 × 2 = 4 km

Second Methdod :
Here, S₁ = 5, t₁ = 6
S₂ = 6, t₂ = 2

Distance =	(S₁ ×S₂)(t₁ + t₂)
	S₂ −S₁

=	(6 × 5)(6 + 2)
	6 − 5

= 30 ×	8	= 4 km
	60