Two trains A and B seperated by a distance of 1200m approach each other at 30m/min.A bird flying at 60 m/min flies from train A to B, returns to A flies back to B and so on till the trains meet.
What is the total distance travelled by the bird?
How many trips between the train does the bird make in all?

• 1200m
  Time taken by trains before meeting = 1200/(30+30)= 20 mins
  Distance travelled by bird during this time = 20*60=1200 mtrs
• 10 years agoHelpfull: Yes(24) No(3)
• A--------1200 m------------B
  A's speed = B's speed =30 m/min
  let A and B meet at point C and both take same time to meet at point C which is t
  so,
  x/30=(1200-x)/30
  =>2x=1200
  x=600 m at point c from A
  so , t=600/30=20min
  in this time bird take total distance is =60*20=1200m
  and total no of turn is one
  
• 10 years agoHelpfull: Yes(10) No(7)
• Meeting time of 2 train 30 + 30 = 1200/t
  t = 20 min;
  Bird traveling time in 20 min = 60*20 =1200 meter and only 1 trip ;
• 10 years agoHelpfull: Yes(5) No(0)
• in first trip
  let t1 is the time taken to cover the distance of 1200m by bird and first train
  30t1+60t1=1200
  t1=40/3
  then,
  distace travelled by first train=30*40/3=400m
  ..........................bird = 60*40/3=800m
  now,remaining=400m
  distance between two trains = 400m
  in second trip bird takes let t2 time
  30*t2+60*t2=400
  t2=40/(3^2) similarly,
  t3=40/(3^3)
  .
  .
  .
  tn=40/(3^n)
  .
  .
  to infinity
  
  total distance travelled by the bird
  =60(t1+t2+t3+........infinite)
  =60[40/(3^1)+40/(3^2)+40/(3^3)+..............]
  =60*40/3[1+1/3+1/9+1/27+................]
  =60*40/3[1/(1-1/3)]
  =1200m
  trips
  the no of trips is infinite ans...
• 10 years agoHelpfull: Yes(4) No(5)
• use relative concept ..
  let train B at rest
  
  BIrd
  AAAA(train)...........B(rest)
  
  speed of train a will be 60 km/hr
  and speed of bird also 60 km/hr.
  after 1st meet distance will be 1200m ;)
  so
  
• 10 years agoHelpfull: Yes(2) No(0)
• total distance=1200
  no. of trips =4
  
• 9 years agoHelpfull: Yes(2) No(0)
• total 1200m will be traveled by bird till both the trains approaches each other..
• 10 years agoHelpfull: Yes(0) No(3)
• let B is in rest then bird's speed wrt B is 90 m/min and A's speed is 30 m/min wrt B . So time when bird reaches train B is 1200/90 , actual distance covered by bird is 60*1200/9=800 ,till then train A covers 30*1200/90=400m , remaining distance=1200-400=800m.
  in 2nd trip let A be in rest then Bird's speed wrt A is 90 & B's speed is 30 wrt A so time taken = 800/90 ,distance covered by bird is 800*60/90=800*2/3 , train covers 30*800/9 ,distance remaining=>800-30*800/90= 800(1-1/3)=800*2/3
  so this will form an GP 800 + 1600/3 + 3200/9 + 6400/27 + 12800/81 ...
  S=800/(1-2/3)=2400 ans.
  No. of trips made by bird are infinite
  
• 10 years agoHelpfull: Yes(0) No(12)
• Time taken by two trains before meeting
  = 1200/(30+30)
  = 20 mins..
  
  Distance traveled by bird during this time
  = 20*60
  =1200 mts..
  
  If we consider Half time...ie., 10min
  Both A & B are 600mts apart by moving 300mts each.. At that moment bird reaches Mid point..
  In similar pattern, After 5more min, Bird reaches train B... In next 5min both trains and bird meets again..
• 10 years agoHelpfull: Yes(0) No(0)
• 
  total distance is 1200m
  time to two train meet each other= 20
  time of first trip=1200/60+30=40/3
  total distance reduced=400 to each train travelled
  distance between two train before second tour=1200-800=400
  time req for second tour =400/90...
  continue it till distance b/w train is approx to 0
  
  total distance travelled=60*20=1200m
• 9 years agoHelpfull: Yes(0) No(0)