A train after travelling 150 km meets with an accident and then proceeds at 3/5 of its former speed and arrives at its destination 8 hours late. Had the accident occurred 360 km further, it would have reached the destination 4 hours late. What is the total distance travelled by the train ?
(a) 960 km
(b) 870 km
(c) 840 km
(d) 1100 km

• ans is b.870km
  let x is disatnce,y is speed, z is time taken when there is no accident
  so
  150/y+(x-150)*5/3y=z+8--------(1)
  510/y+(x-510)*5/3y=z+4---------(2)
  solve (1)&(2) you will get y=60
  so speed of the train is 60 km
  substitute y value in below equations
  150/60+(x-150)*5/3=z+8
  510/60+(x-510)*5/3=z+4
  you will get value of x=870
  so the speed of train is 870km
• 9 years agoHelpfull: Yes(3) No(1)
• sorry hari prasad you are right..
  my previous ans 570 km is coreect IF ACCIDENT OCCURS AT 360KM
  BUT NOW QUESTION IS ACCIDENT OCCURS 360 KM FURTHER
  SO speeds are in ratio of 1:(3/5)=5:3
  so they are 5x and 3x
  d/s=t
  360/3x -360/5x =8-4
  on solving this we get x=12 so actual speed is 5*12=60kmph and reduced speed is 36kmph
  now let take total distance is y km so 150 km with 60 kmph and y-150 with 36 kmph and it takes 8 hours more than actual time
  actual time t= y/60
  150/60 + (y-150)/36 =t+8=y/60+8
  on solving for y we gt y=870km
• 8 years agoHelpfull: Yes(3) No(0)
• speeds are in ratio of 1:3/5 so they are 5x and 3x
  now diff of dist is 360-150=210
  (210/3x) -(210/5x) =(8-4)
  solving this we get x=7
  so actual speed is 35kmph and reduced speed is 21 kmph
  Let total distance be y km
  now 150 km with 35kmph and y-150 km with 21 kmph
  if accident doesnot occur y/35 =t
  150/35 +(y-150)/21 = t+8 =y/35 +8
  570 but am damn sure speed is not 60 kmph just think reaers
• 8 years agoHelpfull: Yes(1) No(1)
• d1 + d2*t1/ t2-t1
  
  150 + 360 * 8/ 8-4
  150 + 360*2
  150+720
  870
• 8 years agoHelpfull: Yes(1) No(0)
• i think it is wrong because while solving for y we are getting -36 not 60
• 8 years agoHelpfull: Yes(0) No(0)