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a car travelled first 50 km at 10 kmph faster than the usual speed but it returned the same distance at 10 kmph slower than the usual speed . if the total time taken by the car is 12 hours, then how many hours does it travel at faster speed?

Read Solution (Total 1)

let the usual speed = x
50/(x+10) is the forward journey time... (i)
50/(x-10) is the return journey time .... (ii)
sum of these is 12 hrs
(i) + (ii) = 12

solving these, we will get the usual speed as 15

faster speed = 15+10 =25

travelling time with faster speed = 50/25 = 2hrs
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