Latest Elitmus Aptitude Question SOLUTION: there are two trains A and B ,A starts from mumbai at 10am @ 51kmph towardsdelhi and B start from Delhi to Mumbai @ 50kmph at 11am distance between them is 273km and each min of

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(#M40041336) ELITMUS Numerical Ability question time and distance Keep an EYE

there are two trains A and B ,A starts from mumbai at 10am @ 51kmph towardsdelhi and B start from Delhi to Mumbai @ 50kmph at 11am
distance between them is 273km and each min of time when both comes towards each other B counts the distance so how many times B calculates distance as whole number.

Asked In Elitmus Rakshit (9 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Pooja Malviya  (9 years ago)

ans: 2 times
speed of train A=51kmph
speed of train B=50kmph
distance between trains=273km
train B starts after 1 hr , at that time train A has already covered 51 km
i.e. remaining distance between train=273-51=222km
relative speed of train=51+50=101kmpr
in min =101/60kmpm
101 is a prime number.
time taken by trains to cover the distance 222km=222/101= approx 2hr20 min
that means after every 60 min the driver will get a whole number
therefore he will get whole number only 2 times.

Is this solution Helpfull? Yes (29) | No (1)  

jhilik roy chowdhury  (9 years ago)

relative spd=101km/p,distance left =(273-51)=222 km..
it will take to cover approx 2.2 hrs.means approx 132 mins. so he counts 132 mins

Is this solution Helpfull? Yes (4) | No (13)  

Saroj kumar  (8 years ago)

ans: 2 times
speed of train A=51kmph
speed of train B=50kmph
distance between trains=273km
train B starts after 1 hr , at that time train A has already covered 51 km
i.e. remaining distance between train=273-51=222km
relative speed of train=51+50=101kmpr
in min =101/60kmpm
101 is a prime number.
time taken by trains to cover the distance 222km=222/101= approx 2hr20 min
that means after every 60 min the driver will get a whole number
therefore he will get whole number only 2 times.

Is this solution Helpfull? Yes (1) | No   

Yeshwanth N S  (9 years ago)

Relative speed=90kmhr, distance b/w them @ 1pm is 222km(273-51)..
Now B calculates distance b/w them in every min so time=222/90 approx time=2.40hr i.e =160min so he counts 160 times.

Is this solution Helpfull? Yes | No (13)  

anshu dubey  (9 years ago)

distance traveled at 1 pm:
A=51km and B=0 (count=0)
at 2 pm:
A=51km and B=49 (count= 60 )
at 3 pm:
A=51 km and B=49(count=60+60)
Distance remaining till 3 pm=273-(51*3+49*2)=20 km
at 3:12 pm:
A=11km(approx) and B=9km(approx) (count=60+60+12)
so B counts till 3:12 pm=132 times

Is this solution Helpfull? Yes | No (3)  

Ashmit Kumar  (9 years ago)

distance left =(273-51)=222 km.
distance travel by both in 1 min=51/60+50/60=101/60=1.6833.
No. of times B calculates distance=222/1.6833=131.8838.
means B will count 131 times.

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Ashmit Kumar  (9 years ago)

distance left =(273-51)=222 km.
distance travel by both in 1 min=51/60+50/60=101/60=1.6833.
No. of times B calculates distance=222/1.6833=131.8838.
means B will count 131 times.

Is this solution Helpfull? Yes | No   

Vaibhav Agrawal  (5 years ago)

Ans:- 1 Time

speed of train A=51kmph
speed of train B=50kmph
distance between trains=273km
train B starts after 1 hr , at that time train A has already covered 51 km
i.e. remaining distance between train=273-51=222km
relative speed of train=51+50=101kmph
time taken by trains to cover the distance 222km=222/101= approx 2hr20 min or 131 min Apprx.

101/60=1.68km/min. or 222/131=1.68km/min

when B count distance at 100th min then the total distance remaining is :-
222-(1.68*100)=54km.

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