A train moving at 3/5^{th} of its normal speed reaches its destination 10 hours late. Find the normal time taken by it.

Let the speed and time be s and t respectively. New speed = 3/5 s and new time = t + 10 According to the question, S × t = 3/5 s(t + 10) St =3/5 st + 6s 5st = 3st + 30s 2st = 30s T = 15 hrs. Alternate Method: Speed ratio = 1: 3/5 = 5:3 So time ratio = 3:5 so here time gap in ratio = 5-3 = 2 ==> 10 1 ==>5 So actual time = 3*5 = 15 hrs

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