Latest Elitmus Aptitude Question SOLUTION: A , B and C are standing on a circular track of 4300m such that A and C are diametrically opposite and B is in between A and C .There clockwise order are ABC , B and C runs clockwi

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(#M40153599) ELITMUS Numerical Ability question Keep an EYE

A , B and C are standing on a circular track of 4300m such that A and C are diametrically opposite and B is in between A and C .There clockwise order are ABC , B and C runs clockwise while A runs anti-clockwise .If the speed of A B and C are 5m/s , 8m/s and 10m/s respectively . They all start running at 10:00 AM. calculate the distance covered by B when A and C meet for the first time after 10:05AM .
I dont remember the options , anyway options cant actually help in such type of questios!!

Asked In Elitmus Ajitesh rai (8 years ago)

Solved Vikas Jha Read Solution (12)

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Read Solution (12) :

Vikas Jha  (8 years ago) Best Solution

Relative speed of A & C will be.. 15 m/s.. hence for first they will meet at
2150/15= 143.33 sec (around 2min 23sec).. so time will be 10:02:23..this case will not be considered as it is before 10:05.. Again they will meet at
4300/15=286.66 sec(around 4min 46sec).. time will be 10:07:10.. this case will be considered
Hence in 7min10 sec i.e 430
In 430sec B will cover 430*8 metres=3440 metres

Is this solution Helpfull? Yes (80) | No   

dipankar saha  (8 years ago)

I think ans is 3440......because in 5 mnt A goes 1500 m and C goes 3000 m....so after 5 mnt distance between A and C is 1950.....after 5 mnt B covers distance 2400 mtr...
relative speed of A and C is 15 m/s....... so time needed to cover the distance 1950/15=130 sec......
in 130 sec B will cover distance 130*8=1040 mtr....so total distance covered by B will be 3440 mtr........

Is this solution Helpfull? Yes (6) | No   

Naveen Kumar Singh  (8 years ago)

I think answer will be approx 3545m.

In the first 5 min from 10:00 to 10:05 B has covered 2400m.
After that A &C took (2 min & 23 sec = 143 sec) to meet. by this time B has covered (143*8 = 1144m).
so total distance covered by B is (2400+1144) = 3544.

I am not sure this option was there in the question or not.

Is this solution Helpfull? Yes (2) | No (5)  

RINKI KESHRI  (8 years ago)

speed of A & C=15m/s
so,2150/15=143.33s
so time=10:02:23
discarded
next they will meet at 4300/15=286.66s=10:07:10 i.e 430 sec
so B will cover 430*8=3440m=3.440km

Is this solution Helpfull? Yes (2) | No   

Lakshmipathi Bojanapu  (8 years ago)

A and C will meet second time at 10:10 Am
hence B covered 80 mts.

Is this solution Helpfull? Yes (1) | No (7)  

chandan kumar  (8 years ago)

since A & C both traveling towards each other so

relative speed =15m/s

when they meet 1st time its between A and C so s=4300/2=> time taken =distance/speed=430/3 < 60*5

=> when they meet 2nd time distance =4300 time taken=860/3< 300
for third time distance=6350 time=6350/15=430 second>300
distance traveled = 8*430=3440

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jagrti porwal  (8 years ago)

How 7.10 Will come... when 4.46 Sec come out after calculation...

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sushmita dey  (8 years ago)

plzz give complete solution

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Naveen Kumar Singh  (8 years ago)

sorry for the previous solution. Answer will be 3440.

After 2 min 10 sec (130 sec) A&C will meet. In the mean while B will cover (8*130 = 1040m)
total distance by B is (2400+1040) =3440m.

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chandan kumar  (8 years ago)

since A & C both traveling towards each other so

relative speed =15m/s

when they meet 1st time its between A and C so s=4300/2=> time taken =distance/speed=430/3 < 60*5

=> when they meet 2nd time distance =4300 time taken=860/3

Is this solution Helpfull? Yes | No (1)  

utkarsh singh  (8 years ago)

860 is the answer.

Is this solution Helpfull? Yes | No (2)  

Neeraj Shukla  (8 years ago)

in all the solution all of you just consider only a and c and wha about b
first take relative speed of b ans that is 2 and thid add to a that gives 7m/s.
now devide this to distance gives first time where they meet. this time when multiply any speed give meeting point

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