1. A train leaves a station A at 7 am and reaches another station B at 11 am. Another train leaves B at 8 am and reaches A at 11.30 am. The two trains cross one another at





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  • By: anil on 05 May 2019 01.53 am
    Time taken by 1st train to travel from A to B = 11-7 = 4 hours
    Time taken by 2nd train to travel from B to A = 11:30-8 = 3.5 hours => ratio of time taken by 1st train to 2nd train = $$4 : frac{7}{2}$$ = 8 : 7 Since, speed is inversely proportion to time => Ratio of speeds of 1st train to 2nd train = 7 : 8 Let the speed of 1st train = $$7x$$ and 2nd train = $$8x$$ km/hr
    Distance between the two stations = time * speed = $$7x * 4 = 28x$$ km
    We know that, 1st train starts one hour early, thus it will cover $$7x$$ distance till the time 2nd train starts. So, at 8.00 a.m., remaining distance between two trains = $$28x-7x = 21x$$ km Also, the two trains are moving in opposite directions, =>relative speed of two trains = $$8x+7x = 15$$ km/hr Now, time taken to meet = $$frac{21x}{15x} = frac{7}{5}$$ hours => $$frac{7}{5} * 60$$ = 84 minutes after 8.00 a.m. => Time when they meet = 8.00 a.m. + 84 min = 9:24 a.m.
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