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Two trains are moving in the same directions at 65 km/hr and 45 km/hr. The faster train crosses a man in slower train in 18 seconds. The length of the faster train is
A. 100 m
B. 120 m
C. 145 m
D. 180 m

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Two trains are moving in the same directions at 65 km/hr and 45 km/hr. The faster train crosses a man in slower train in 18 seconds. The length of the faster train is
A. 100 m
B. 120 m
C. 145 m
D. 180 m
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20 Answers

  1. Relative speed =65-45km/hr
    Relative speed=20km/hr
    Convert to m/s
    S=20*5/18
    S=5.555
    Length of the faster train =5.55555*18seconds
    =100meters

  2. A man in slower train ;speed =45 km/h
    Faster train speed = 65 km/h
    crossed speed is =difference of speed = 65km/h – 45 km/h = 20 km/h
    Length of train = (20 *18)/3600=0.1 km = 0.1*1000=100 meter

  3. 65 faster train
    45 slower train
    faster train crosses slower one by 20 kmph
    and if it crosses the slower one in 18 sec then the length of faster train is
    20*1000*18/3600
    100m
    so the length of the faster train is 100m

  4. Assuming 1st train length is = X mtr.
    Relative speed of both train will be = (65-45)*5/18 = 100/18 Km/Hrs.
    1st train taking time to crossing the boy sitting in slower train is = 18 sec.
    Then equation will be :
    Distance = (Relative Speed) * (total taken Time)
    X = (20*5/18) * (18)
    Then Train distance (X) = 100 Mtrs.

  5. relative speed is=65-45=20kmph, 20*(5/18)=5.5555m/s
    time=18sec
    distance=speed*time=5.5555*18=100m

  6. Total Distance =?

    Time = 18

    20 = (X/18) x (18/5) => 100

  7. 65-45km/h=20km/h
    D=s*t
    therefore 20km/h * 1000 =20000m * 18/3600
    20000 * 1/200=100m

  8. 65 – 45 = 20 kmph
    20 * 5/18 =50/9 m/s
    50/9* 18 = 100m/s

  9. relative speed is=65-45=20kmph, 20*(5/18)=5.5555m/s
    time=18sec
    distance=speed*time=5.5555*18=100m