Abus went halfway from a village to a city at a certain speed. then it stopped in traffic for an hour. in order to catch up the time, for the remaining 200 km of the way the driver decided to go at a speed which was 10 km/h greater than his speed before he got stuck in traffic. the bus got to the city on time. what was the speed of the bus before the traffic holdup?

We have to assume that the speed before being stuck was sufficient to get to the destination on time had there been no delay. Call that speed "s" in km/h.

Since 200 km is "halfway", the total distance must be 400 km.

time = distance / speed
total time = (time for first half) + (delay) + (time for second half)

400/s = 200/s + 1 + 200/(s+10) . . . .times are in hours, distances in km
200/s = 1 + 200/(s+10) . . . . . . . . . . subtract 200/s
200(s+10) = s(s+10) +200s . . . . . . .multiply by s(s+10)
0 = s² +10s - 2000 . . . . . . . . . . . . . .subtract the left side
(s+50)(s-40) = 0
Solutions are s = -50, s = 40

The speed of the bus before the traffic holdup was 40 km/h.

The slope best fit isc. 1/6 mile per hour