Question

A bus leaves the stop 30 minutes before the scheduled time. the driver decreases its speed by 30km/hr. at the next bus stop 180 km away, the bus reached on time. find the original speed of the bus?

Answer 1

Let the original speed be 'S' km/hr
180/S = T hrs
The bus leaves 30 minutes early, so at usual speed it might have reached 30 minutes early. But the driver reduced the sped by 30 km/hr so as to reach the destination (180 km) on scheduled time.
So, if the speed was 'S'
180/S = T - 30/60 hrs....(i)
Speed decreased by 30
180/S-30 = T.....(ii)
(ii) - (i)
180/S-30 - 180/S = 30/60
180[1/S-30 - 1/S] = 1/2
180[S - (S-30)]/S*(S-30) = 1/2
360*30 = S*(S-30)
120*3*30 = S*(S-30)..... [360 can be written as: 120*3]
120*90 = S*(S-30)
Now look at L.H.S & R.H.S
If S = 120, S-30 =90
Hence, S = 120... [Alternatively, quadratic equation solving can also be done to reach this answer]
Hence, original speed of bus = 120 km/hr
Hope it helps.