Two trains for Palwal leave Kanpur at 10 a.m and 10:30 am and travel at the speeds of 60 Kmph and 75 Kmph respectively. After how many kilometers from Kanpur will the two trains be together?

• Let two train meet at X km distance.
  Time taken by first train to cover X km=x/75 h
  Time taken by 2nd train to cover X km=x/60 h
  Now, x/60 - x/75=1/2
  thus, x=150.
• 8 years agoHelpfull: Yes(80) No(6)
• If two trains leave say 'P' for 'Q' at time 't1' 't2' and with speeds 'a' and 'b' ,
  Then the distance D from p where they meet, is given by
  D = diff in time (t2-t1) X a*b / b-a
  There u go ...
• 9 years agoHelpfull: Yes(33) No(18)
• relative speed =75-60=15kmph
  in half an hour 1 train travelled 30km.
  to reach I st train 2 nd train need 30/15=2 hours.
  in two hours 2 nd train travelled 75*2=150km
• 9 years agoHelpfull: Yes(23) No(3)
• K----------------------------------------------------------P
  T11 10:00 60 km/h
  TT12 10:30 75 km/h
  Suppose they meet after t time from 10.00
  75(t- 1/2)=t *60
  t=2.5
  Distance travelled till meeting=75*(2.5-.5)=150 km
• 9 years agoHelpfull: Yes(13) No(8)
• As we know that Train A and Train B leaves from the same station ,
  simple trick for solving that is find the factors of both .
  5 |60 , 75
  ------------------
  3 | 12 , 15
  ------------------
  4 | 4 , 5
  -----------------
  5 | 1 , 1
  
  (5*3*4*5 ) / 2= 150
• 7 years agoHelpfull: Yes(9) No(3)
• travelled distance is same so
  60*t=75*(t-1/2)
  t=5/2
  s=60*5/2=150
• 6 years agoHelpfull: Yes(5) No(0)
• x+1x=3
  
  ⟹(x+1x)2=9
  
  ⟹x2+1x2+2=9
  
  ⟹x2+1x2=7
  
  Again,
  
  (x+1x)3=27
  
  ⟹x3+1x3+3(x+1x)=27
  
  ⟹x3+1x3=27−3×3=18
  
  Now,
  
  (x2+1x2)(x3+1x3)=(x5+1x5)+(x+1x)
  
  ⟹18×7=x5+1x5+3
  
  ⟹x5+1x5=18×7−3=123
• 8 years agoHelpfull: Yes(0) No(16)
• 2.5 dm = .0.25m
  19.25 m × 2 m = 38.5 m^2
  12.50 m × 2 m = 25 m^2
  Since they will overlap therefore we have to subtract 2×2 m^2
  Therefore answer will be
  (25 + 38.5 - 4 ) × 1.32
  I.e 78.65
• 6 years agoHelpfull: Yes(0) No(2)
• Solving logically .
  When train A = 30
  Train B = 0
  
  Train A = 31
  Train B = 1.25
  
  Train A = 32
  Train B = 2.5
  
  Train A = 34
  Train B = 5
  
  Similarly @ a at 38 and B at 10
  
  Now if we subtract 5 from 34 that will be 29.
  The 10 from 38 ie 28...and this will continue. When it will be 0? After 30 steps.
  Therefore 5 × 30 will come 150. Or 30 (train A)+ (30×4(common d)) will also give 150.
• 6 years agoHelpfull: Yes(0) No(0)
• First train start in 10 am and 2nd train start train 10:30
  So the 1st train cover 30 km distance in 10:30.
  Now, both train time are equal
  Let the train distance = x
  time 1= time 2
  x-30/60=x/75
  So x=150 km
• 5 years agoHelpfull: Yes(0) No(0)
• A-10am
  B-10:30 am
  Speed(A)=60kmph So when B starts A would have covered 30km and when they meet they will have same time. let x be the distance covered by b when they meet.
  (30+x)/60=x/75
  x=150 km.
• 5 years agoHelpfull: Yes(0) No(0)
• S1*S2*t / S1-S2
  (60*75*1/2 )/60-75
  (60*75*1/2)/15
  60*5*1/2
  60*2.5
  150.0km
• 5 years agoHelpfull: Yes(0) No(0)