Latest Elitmus Aptitude Question SOLUTION: Two trains starts to run with a speed of 5kmph and 7kmph from two stations A & B respectively towards each other. Both the trains reverse its direction after reaching either of the

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(#M40033156) ELITMUS Numerical Ability question Elitmus 23rd feb14 Keep an EYE

Two trains starts to run with a speed of 5kmph and 7kmph from two stations A & B respectively towards each other. Both the trains reverse its direction after reaching either of the station. The distance between the two stations is 27km.If they both starts to run at 06:00am, at what time they will meet exactly in between the two stations for the second time?

Asked In Elitmus Vivek Jain (10 years ago)

Unsolved Read Solution (16)

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Read Solution (16) :

Sourav Chandra  (10 years ago)

distance between them is 27 km,a goes 7km in 1 hr and b goes 5 km in 1 hr...
now in 6 hrs b reaches a and returns a distance of 3km in reverse direction whereas a reaches b and returns in reverse direction of 15km... so distance left between them is (27-(15+3))=9km... let both the trains meet 9 km in t hours... therefore 5t+7t=9 or t=45 mins... thus they will meet for the second tym at 12:45pm.

Is this solution Helpfull? Yes (34) | No (11)  

PRAVIN KUMAR  (9 years ago)

For 2nd meeting total distance covered=3d=3*27
relative speed=5+7=12
so time taken for 2nd meet=27*3/12=6hr45min
so they meet 2nd time at 12:45pm

Is this solution Helpfull? Yes (17) | No (1)  

Anurag Yadav  (9 years ago)

Fr meeting 2nd time they cober total distance of 27*3=81. Now (7+5)×6hr=72 km..remaining distance (81-72) and time taken=9*60/12=45 min hence 6+6.45=12.45

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Nancy neha  (10 years ago)

1st meet is after 27/12hr=2.25hr
therefore 2nd meet will be after 2.25*2=4.5hr
the 2 trains will meet at 10:30am for the 2nd tym....

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anupam jaiswal  (9 years ago)

ans should be 5.4+1.34=6.74 hrs

Is this solution Helpfull? Yes (2) | No   

Rahul Sachan  (10 years ago)

for first time: on applying relative motion 27/(5+7)=2.25
for second time : 13.5/5 + 13.5/5 = 4.6
total time = 6.85 hrs
hence they will meat on 12:51

Is this solution Helpfull? Yes (1) | No (3)  

Debabrata Nayak  (10 years ago)

ans:time=10:30am
7x+5x=27+27=>x=9/2=4hrs 30 minutes where x is the time after 6 am when they met second time.here distance covered by faster train 27+27/2 and slower train=27/2

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Mukesh Nayal  (10 years ago)

at 12:48pm

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Mani kandan  (10 years ago)

pls share me correct solutions

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harsh  (9 years ago)

correct answer is after 395 minutes

Is this solution Helpfull? Yes | No (1)  

Prakash Singh  (9 years ago)

i think answer will be 6:00 A.M + 6.74hr

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Pankaj Kumar  (9 years ago)

Suppose A and B two trains in first meeting A covers x km so B covers (27-x)
and time will be constant .
So by using formula
5/x = 7/(27-x) solve dis
distance x=11.25 ; Suppose similar 2nd meeting B cover y km so A again cover (27-y)km after reach End points ..
Mean B cover's 11.25 + y
A cover's (27-11.25)+ (27-y)= 42.75-y
so again take formula here time constant
(11.25+y)/7 = (42.75-y)/5
y=29.625
so Now we will calculate time taken by first meeting and second meeting with respect to A (you can also take w.r.t B)
first meeting by A at time t1= 11.25/5 = 2.25 hrs and
second meeting by A at time t2= (42.75-29.625)/5 = 2.625 hrs
so total time t1+t2= 4.875 hr mean we can say (4 hr 87 min) mean (5 hr 27 min)
so time will be 5.27 hr
train start at 6:00 AM so it will take 11:27 AM

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ABHISHEK GHOSH  (9 years ago)

let the train meet second time at x distance.

1st train travel from station A to B = 27 km
after reversing the direction first train travel from station B to point where it meets i.e =(27-x) km
trotal distance travel by first train=27+(27-x)=54-x
time = 54-x/5 ------1

2nd train travel from station B to A = 27 km
after reversing the direction 2nd train travel from station A to point where it meets i.e =x km
trotal distance travel by first train=27+x
time = 27+x/7 -----2

equating 1 & 2 (time constant)

54-x/5=27+x/7
x=243/12

putting x value in eq 2
t=(27+(243/12))/7
=6.75 hr

6+6.75= 1:15 pm

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abhinav pandey  (8 years ago)

time taken by first train to travel to its destination=5.4 hrs
time taken by second train to travel to its destination=3.4 hrs(approx.)
difference between the time taken by the trains to reach their respective destinations=(5.4-3.4)=2hrs
for 2hrs the distance travelled by the second train having speed 7 kmph=14km
hence the time taken by the trains to meet in the reverse journey=(27-14)/12=1.08hrs(approx..)
total time taken by the trains to meet=5.4+1.08=6.48hrs
hence ans is 12.48 pm

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sanjay kumar gupta  (8 years ago)

slower 1 take 5 hr 24min...in ds tym the other 1 cover 16.15 km dis. now the dis. between dem 10.85km....there fore result--12:18pm

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rahul kumar  (7 years ago)

time total distance traveled by both of the trains to meet 2nd time= 3*d(using formula "n*d", here n=nthtime meet)|,
=>total =3*27=81.
=> time=81/relative speed
=> 81/12
=>6.75 hrs
=>6+6.75 hrs
=>1:15 pm(ans)|

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