Latest TCS Aptitude Question SOLUTION: 3) In a race of 4000m around a circular ground of circum 1000m The fastest and slowest athletes meets first time 5 min after the starting of race . the speed of the all athletes re

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3) In a race of 4000m around a circular ground of circum 1000m The fastest and slowest athletes meets first time 5 min after the starting of race . the speed of the all athletes remains constants over the race . The speed of the fatest is double of the slowest athletes. Than in how much time the athletes who wins the race will complete the race.......

Asked In TCS Ankush (11 years ago)

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AYUSH SRIVASTAVA  (11 years ago)

take the speed of slowest to be = s m/min
then speed of fastest will be = 2s m/min
now we know that slowest moves x in 5 min
so,
x/s=5
i.e x=5s
now in case of fastest he travels 1000+x m to meet the slowest one
so (1000+x)/2s=5
as x=5s,
we solve 1000+5s=10s
i.e s=200 m/min for the slowest nd for fastest it is 2s i.e 400 m/min
now total time = 4000/400 = 10 min

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Arka Laha  (11 years ago)

fastest runner will always win the race let speed of him is 2x and slowest runner speed x,so from que 10x-1000=5x as they meet after 5 min so x=200 fastest runner speed is 400 so after 5 min he will go 2000 and slowest one will go 1000 they meet at starting point so ans is 10 am i correct??

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