Let the speed of the train during returning journey be x km/h
Speed during onward journey = x + 25x/100 = 5x/4 km/h.
Distance coverd in onward journey = 800 / 2 = 400 km
Total time taken = Covered distance / Speed
Time taken by train in onward journey = 400/(5x/4)
and time taken in returning journey = 400/x
Thus, 400/(5x/4) + 400/x = 16
? 320/x + 400/x = 16
? 16x = 720
? x = 45 km/h
Speed of the train in the onward journey = 5 x 45/4 = 56.25 km/h
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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