Two trains starts to run with a speed of 5kmph and 7kmph from two stations A & B respectively towards each other. Both the trains reverse its direction after reaching either of the station. The distance between the two stations is 27km.If they both starts to run at 06:00am, at what time they will meet exactly in between the two stations for the second time?

• distance between them is 27 km,a goes 7km in 1 hr and b goes 5 km in 1 hr...
  now in 6 hrs b reaches a and returns a distance of 3km in reverse direction whereas a reaches b and returns in reverse direction of 15km... so distance left between them is (27-(15+3))=9km... let both the trains meet 9 km in t hours... therefore 5t+7t=9 or t=45 mins... thus they will meet for the second tym at 12:45pm.
• 10 years agoHelpfull: Yes(34) No(11)
• For 2nd meeting total distance covered=3d=3*27
  relative speed=5+7=12
  so time taken for 2nd meet=27*3/12=6hr45min
  so they meet 2nd time at 12:45pm
• 9 years agoHelpfull: Yes(17) No(1)
• Fr meeting 2nd time they cober total distance of 27*3=81. Now (7+5)×6hr=72 km..remaining distance (81-72) and time taken=9*60/12=45 min hence 6+6.45=12.45
• 9 years agoHelpfull: Yes(6) No(0)
• 1st meet is after 27/12hr=2.25hr
  therefore 2nd meet will be after 2.25*2=4.5hr
  the 2 trains will meet at 10:30am for the 2nd tym....
• 10 years agoHelpfull: Yes(4) No(10)
• ans should be 5.4+1.34=6.74 hrs
  
• 9 years agoHelpfull: Yes(2) No(0)
• for first time: on applying relative motion 27/(5+7)=2.25
  for second time : 13.5/5 + 13.5/5 = 4.6
  total time = 6.85 hrs
  hence they will meat on 12:51
• 10 years agoHelpfull: Yes(1) No(3)
• ans:time=10:30am
  7x+5x=27+27=>x=9/2=4hrs 30 minutes where x is the time after 6 am when they met second time.here distance covered by faster train 27+27/2 and slower train=27/2
• 10 years agoHelpfull: Yes(1) No(3)
• at 12:48pm
• 10 years agoHelpfull: Yes(0) No(1)
• pls share me correct solutions
• 10 years agoHelpfull: Yes(0) No(0)
• correct answer is after 395 minutes
• 10 years agoHelpfull: Yes(0) No(1)
• i think answer will be 6:00 A.M + 6.74hr
• 9 years agoHelpfull: Yes(0) No(0)
• Suppose A and B two trains in first meeting A covers x km so B covers (27-x)
  and time will be constant .
  So by using formula
  5/x = 7/(27-x) solve dis
  distance x=11.25 ; Suppose similar 2nd meeting B cover y km so A again cover (27-y)km after reach End points ..
  Mean B cover's 11.25 + y
  A cover's (27-11.25)+ (27-y)= 42.75-y
  so again take formula here time constant
  (11.25+y)/7 = (42.75-y)/5
  y=29.625
  so Now we will calculate time taken by first meeting and second meeting with respect to A (you can also take w.r.t B)
  first meeting by A at time t1= 11.25/5 = 2.25 hrs and
  second meeting by A at time t2= (42.75-29.625)/5 = 2.625 hrs
  so total time t1+t2= 4.875 hr mean we can say (4 hr 87 min) mean (5 hr 27 min)
  so time will be 5.27 hr
  train start at 6:00 AM so it will take 11:27 AM
• 9 years agoHelpfull: Yes(0) No(1)
• let the train meet second time at x distance.
  
  1st train travel from station A to B = 27 km
  after reversing the direction first train travel from station B to point where it meets i.e =(27-x) km
  trotal distance travel by first train=27+(27-x)=54-x
  time = 54-x/5 ------1
  
  2nd train travel from station B to A = 27 km
  after reversing the direction 2nd train travel from station A to point where it meets i.e =x km
  trotal distance travel by first train=27+x
  time = 27+x/7 -----2
  
  equating 1 & 2 (time constant)
  
  54-x/5=27+x/7
  x=243/12
  
  putting x value in eq 2
  t=(27+(243/12))/7
  =6.75 hr
  
  6+6.75= 1:15 pm
  
• 9 years agoHelpfull: Yes(0) No(4)
• time taken by first train to travel to its destination=5.4 hrs
  time taken by second train to travel to its destination=3.4 hrs(approx.)
  difference between the time taken by the trains to reach their respective destinations=(5.4-3.4)=2hrs
  for 2hrs the distance travelled by the second train having speed 7 kmph=14km
  hence the time taken by the trains to meet in the reverse journey=(27-14)/12=1.08hrs(approx..)
  total time taken by the trains to meet=5.4+1.08=6.48hrs
  hence ans is 12.48 pm
• 8 years agoHelpfull: Yes(0) No(1)
• slower 1 take 5 hr 24min...in ds tym the other 1 cover 16.15 km dis. now the dis. between dem 10.85km....there fore result--12:18pm
  
• 8 years agoHelpfull: Yes(0) No(1)
• time total distance traveled by both of the trains to meet 2nd time= 3*d(using formula "n*d", here n=nthtime meet)|,
  =>total =3*27=81.
  => time=81/relative speed
  => 81/12
  =>6.75 hrs
  =>6+6.75 hrs
  =>1:15 pm(ans)|
  
• 7 years agoHelpfull: Yes(0) No(0)