The distance is constant in this case.
Let the time taken for travel with a speed of 10 kmph be 't'.
Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.
Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)
Extra time taken = t - 2t/3 = t/3
=> 1pm - 11am = 2hrs
=> t/3 = 2h
=> t = 6 hrs.
Now, Distance = speed x time = 10 x 6 = 60 kms
Time he takes to reach at noon = 6 - 1 = 5 hrs
Now, Speed = 60/5 = 12 kmph.
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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