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.
we know that speed = distance traveled/time taken
let the total distance traveled by the car is 2x km.
then time taken by it to cover first half is x/60 hour.
and for second half is x/40 hour.
Then average speed= total distance travelled / total time taken.
i.e.
=>
= 48 kmph.
Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.
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.
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
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