Latest Elitmus Aptitude Question SOLUTION: A train 300 m log overtook a man walking along the line (in the same direction of the train) at the speed of 5km/hr and passed him in 30 sec. the train reached the station in 15 mi

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(#M40032429) ELITMUS Numerical Ability question Keep an EYE

A train 300 m log overtook a man walking along the line (in the same direction of the train) at the speed of 5km/hr and passed him in 30 sec. the train reached the station in 15 min it hass passed the man . in what time man reach the station....

Asked In Elitmus ranjeet singh (10 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

RAKESH  (10 years ago)

speed of the train relative to man =(300/30)m/sec =10 m/sec =10*18/5 = 36 km/h

let the speed of the train be x km/hr
then, relative speed =(x-5) km/hr

so, (x-5)=36 => x=41 km/h

train reached the station in 15 min
so distance travelled by train in 15 min = 41*15/60= 10.25 km
time taken by man = 10.25/5 = 2.05 hr =123 min

Is this solution Helpfull? Yes (66) | No (3)  

mahima srivastava  (10 years ago)

distance covered by man in 30 sec= 30*5*5/18=125/18 m
distance covered by train to overtake the man= 300+distance covered by man = 300+125/18 = 1025/3
speed of train= (1025/3)/30 = 205/18 m/s
distance between man and station (after overtake)= (205/18)*15*16= 10250 m
time taken by man= 10250 *18/5*5 =7380 sec = 123 min

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

D.HARINATH REDDY  (10 years ago)

man traveled in 30 sec = 5*30/3600 = 0.0416667km
train traveled to overtook the man = 0.3+.041667 km=.3416667
train speed = .3416667*3600/30= 41kmph
station distance from man when train overtook the man = 41*15/60 =10.25km
time required to man to reach the station =10.25/5 = 2h 3min.

Is this solution Helpfull? Yes (3) | No   

Vaishak Damodar  (9 years ago)

Guys!! This is important. "Rakesh" above have done everything correct until the last step. I am going to correct it;
"speed of the train relative to man =(300/30)m/sec =10 m/sec =10*18/5 = 36 km/h

let the speed of the train be x km/hr
then, relative speed =(x-5) km/hr

so, (x-5)=36 => x=41 km/h

train reached the station in 15 min
so distance travelled by train in 15 min = 41*15/60= 10.25 km
time taken by man = 10.25/5 = 2.05 hr =123 min"

My Correction: Distance travelled by train in 15 min = 10.25km
Now in those 15min, the man is constantly walking too. So, in those 15min, he covers 1.25km. So, he totally has to travel 10.25-1.25 = 9km (After the train stops).
He can do that in time t = 9/5 = 1 hour 48 minutes!!!

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ajay negi  (10 years ago)

distance covered by man in 30 sec= 30*5*5/18 = 125/3 m
distance covered by train in 30 sec = (300+ 125/3)m
and relative speed in same direction = (S-5*5/18) m/s (where S = train's speed in m/s )
so equation of the same direction is :30 = (300+ 125/3)/(S-5*5/18)
S = 115/9 m/s
now train reached the station in 15 min
so distance covered by train in 15 min= 15*60*115/9 = 11500 m
ans time taken by man = 11500/(5*5/18)
= 8280 sec = 138 min.

Is this solution Helpfull? Yes (1) | No (9)  

Prakash Singh  (9 years ago)

I think 93min

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sneha  (9 years ago)

vaishak damodar how 1.25km came,acrdng to ur crrctn wn d man is constantly walking.

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