Inspector s 228 meter behind the thief and now aftersome x distance he will catch the thief. So
x/30 = 228+x/42 => x = 570 m
(228 + 570) / 42 = 19 min
Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time
so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.
Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs
Hence, if bus runs at its own speed, the time saved = 22 - 19.8 = 2.2 hrs.
After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km Without servicing, speed of car = 50 kmph => Required time = Distance/speed = 360/50 = 7.2 hrs.
Avg speed = Tot dist/Tot time = (45+64+75)/23 = 184/23
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