A train cross a 450 m bridge in 42 s and another bridge 650 m length in 50 s what is approximate speed of train in km/h?

• First u hv to find length of train. assume L
  (450 + L)/42 = (650+L)/50
  L= 600m
  Now speed = (450+600)/42 or (650+600)/50 = 25 m/s
  now 25*(18/5) = 90Km/hr
• 10 years agoHelpfull: Yes(44) No(0)
• train is covering 200 m in 8 sec,
  now,go from options 90km/hour=90*5/18=25m/sec
  200/25=8 sec,so hence it should be 90km/h
  
  
• 10 years agoHelpfull: Yes(19) No(5)
• v = 650/50*18/5 km/h
  46.8
  or
  V = 450/43*18/5
  46.8 km/h
  
• 10 years agoHelpfull: Yes(3) No(4)
• 90km/hr
  because 650-450/50-42=200meter/8sec=150meters/6sec=1.5km/min=90km/hr :-)
  
• 10 years agoHelpfull: Yes(3) No(0)
• 90 km/hour
• 10 years agoHelpfull: Yes(2) No(2)
• when considering the train crossing the bridge, the length of the train and bridge gets added up so
  let us consider the lenght of the train in meters
  s1=x+450/42 and s2=x+650/50 as speed of train is same so s1=s2 equate and calculate the ans
  
  hope this is correct
• 10 years agoHelpfull: Yes(1) No(0)
• 90 km/h
  l=600m
  distance=1050
  time=42
  speed=d/t
  (1050/42)*18/5
  =90km/h
• 10 years agoHelpfull: Yes(1) No(0)
• train is covering 200 m in 8 sec
  200/8=25m/s
  25*(18/5)=90km/h
• 9 years agoHelpfull: Yes(1) No(0)
• s=d/t; so s1=450/42=10.71; s2=650/50=13;
  A s=10.71+13/2=12m/s*18/5=43.2km/h;
  
• 10 years agoHelpfull: Yes(0) No(2)
• Vt=(Lt+Lp1)/Tp1----1
  or Vt=(Lt+Lp2)/Tp2----2
  (Lt+450)/42=(Lt+650)/50
  solving this Lt=600m
  putting Lt=600 and Lp2=650 in equation 2
  we get Vt=25m/s
  
• 10 years agoHelpfull: Yes(0) No(0)
• if length of train is l then
  l + 450 traveled in 42 s
  l + 650 traveled in 50 s
  then 200 in 8 sec
  (200/8) * 18/5 = 90
• 10 years agoHelpfull: Yes(0) No(0)
• Difference of Length /Difference in Time = 650-450/50-42
  => 200/8 m/s
  => 25 m/s
  
  now 25*(185)
  
  =>5*18 = 90 Km/hr
• 10 years agoHelpfull: Yes(0) No(0)