Latest Elitmus Aptitude Question SOLUTION: There are two stations Rampur and Kanpur. Distance between them is 273KM. Ist Train starts from Rampur & Travel with the speed of 51KMPH at 1:00 PM and 2nd Train starts from Kanpur

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(#M40037453) ELITMUS Numerical Ability question Keep an EYE

There are two stations Rampur and Kanpur. Distance between them is 273KM. Ist Train starts from Rampur & Travel with the speed of 51KMPH at 1:00 PM and 2nd Train starts from Kanpur & Travel with the speed of 60KMPH at 2:00 PM. 2nd Train's Driver note the distance btw both train after every 1 Minute. Find the no. of times he will get the distance in whole number.
Options:
A) 13
B) 7
C) 5
D)30

Asked In Elitmus Ankit Gupta (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Ankit Chachra  (10 years ago)

At 2 PM: 1st train would have covered 51kms...and 2nd would have just started...

Therefore, at 2 PM
Distance between them: 273-51= 222 Kms
and relative speed: 51+60 = 111 Km/hr

Now after each minute, distance between them will decrease by: 111/60 kms or 37/20 kms

Hence, we need whole number, therefore, denominator 20 should get cancelled...
therefore, we will get whole number after every 20 minutes:

Like: 2.20, 2.40, 3.00, 3.20 and 3.40...TOTAL: 5 TIMES. (at 4 pm, train will meet so no need)

therefore, C) 5 is the answer

Is this solution Helpfull? Yes (62) | No (5)  

anshi  (10 years ago)

the answer is 7.
also, consider the case when t=0 and t=120 mins (t=120 mins wen the 2 trains meet, distance btwn dem is 0.)

Is this solution Helpfull? Yes (15) | No (3)  

Inder  (10 years ago)

5

after every 20 mins, the distance will be decreased by 37 kms as
relative speed = 51+60=111 km/hr = 111/60 km/min= 37/20 km/min.
Trains will meet at 4.00 PM i.e. after 2 hr journey by train from kanpur.

so distance in integers ( KMS) will be at 2.20 PM, 02.40 PM, 03.00 PM, 03.20 PM and 03:40 PM.

Is this solution Helpfull? Yes (7) | No (4)  

yagyesh  (9 years ago)

since the 2nd train started at 2pm, till that time 1st train covers distance 51 kms
hence remaining distance is 273-51=222 kms
we know when two trains are in opposite directions their speed are added hence total relative speed is 60+51 = 111 kms/hr or 1.85 kms/min
now both trains will meet after time 222/111 = 2 hrs..or 120 min
now let distance traveled by both trains till x th minute = 1.85*x
now 222 - 1.85*x = should be a whole no
x=0,20,40,60,80,100,120 satisfies,
at 120th minute distance between them is 0(whole no.)
so the answer will be 7

Is this solution Helpfull? Yes (7) | No (2)  

Ayush Singh  (10 years ago)

its (C) 7
Consider case for t=0 and t=120 also.

Is this solution Helpfull? Yes (6) | No (1)  

JOYDEEP SUR  (10 years ago)

is 0 whole no?

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

Ashwinee Kumar  (9 years ago)

let train starts form Rampur at time x ,
51x+60(x-1)=273
111x=333
x=3h
now, 60/3 min =20 min
so at every 20 min(20*1 min) distance is noted i.e 60/20= 5

Is this solution Helpfull? Yes (5) | No (3)  

nirmal  (10 years ago)

5 is the answer i guess

Is this solution Helpfull? Yes (3) | No (7)  

Ankit Gupta  (10 years ago)

Actually Option B was not there So the Correct Ans is 5.

Is this solution Helpfull? Yes (3) | No   

Arun kumar  (8 years ago)

answr would be 7 b'coz 0 is a whole no.
and @ sharp 2 the distance is 222 which is a whole no.s and we will get other whole nums value of dist. at every 20 mints including @ 4:00.
B'coz 0 is a whole numb.

Is this solution Helpfull? Yes (1) | No   

shashank singh  (8 years ago)

in 1 hr rampur train will cover 51 Km and after that kanpur train starts.
relative distance=273-51= 222 km
relative speed=51+60=111 km/h
in 1 min relative distance=111/60=37/20
to be whole no it should be multiple of 20 i.e after every 20 min it become whole no
2.20,2.40,3.00,3.20.3.40= 5 times

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sahil julka  (7 years ago)

actually it is not given till what time the driver notes the distance between the trains...it might be after they meet and start going away from each other???

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madhuri  (7 years ago)

let total distance of the pool is d.
case 1;
time taken by east diver-50/Se.
time taken by west diver for 1st meeting=(d-50)/Sw.
therefore for 1st meeting point-- 50/Se=(d-50)/Sw.
Se/Sw=50/(d-50)------------------------1

case 2;
time taken by east diver for 2nd meeting= (d-50+20)/Se
time taken by west diver for 2nd meeting= 100+(d-50-20)/Sw
therefore for 2nd meeting=Se/Sw=(d-30)/(d+50)---------2

from eqn 1 and 2 we get
(d-30)/(d+30)=50/(d-50)
solving we get
d=130.
this is the correct sol.

Is this solution Helpfull? Yes | No (1)  

virender kumar  (6 years ago)

The distance travel by the 1st train (started at 1:00 PM) during 1st hour of its journey is 51km.

The second trains starts at 2:00PM and the distance between both the train at that moment is (273-51) = 222KM which must be noted by the driver.

They will meet in 2 hours i,e 4:00 PM (Train 1 will cover 102KM and Train 2 will cover 120KM).

Since the speed of train 2 is 60 KMPH i.e it will travel 1 KM in every 1 Min and for the distance between 2 trains to be a whole nos the minimum time interval which should be taken is 20 min in which train 1 will cover a distance of 17km. So after interval of every 20 min the distance will be a whole nos.

So the times are: 2;00PM, 2:20PM, 2;40PM, 3:00PM, 3;20PM, 3:40PM,4:00PM i.e 7 times.

Here 4:00PM is included because at 4:00PM the distance between them is '0' which is also a whole nos.

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