A bus left 60 minutes later than the scheduled time but in order to reach its destination 48 km away in time, it had to increase the speed by 4 km/hr from the usual speed. What is usual speed (in km/hr) of the bus ?

Usual speed of bus = x kmph.

According to the question,

Time $=frac{ ext { Distance }}{ ext { Speed }}$

$Rightarrow frac{48}{x}-frac{48}{x+4}=1$

$Rightarrow 48left(frac{x+4-x}{x(x+4)} ight)=1$

⇒ x (x + 4) = 48 × 4

= 12 × 16

⇒ x (x + 4) = 12 (12 + 4)

⇒ x = 12 kmph.