There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.

• It is given that the platoon and the last person moved with
  uniform speed. Also, they both moved for the identical
  amount of time. Hence, the ratio of the distance they
  covered - while person moving forward and backword - are
  equal.
  
  Let's assume that when the last person reached the first
  person, the platoon moved X meters forward.
  
  Thus, while moving forward the last person moved (50+X)
  meters whereas the platoon moved X meters.
  
  Similarly, while moving back the last person moved [50-(50-
  X)] X meters whereas the platoon moved (50-X) meters.
  
  Now, as the ratios are equal,
  (50+X)/X = X/(50-X)
  (50+X)*(50-X) = X*X
  
  Solving, X=35.355 meters
  
  Thus, total distance covered by the last person
  = (50+X) + X
  = 2*X + 50
  = 2*(35.355) + 50
  = 120.71 meters
  
  
  Note that at first glance, one might think that the total
  distance covered by the last person is 100 meters, as he
  ran the total lenght of the platoon (50 meters) twice.
  TRUE, but that's the relative distance covered by the last
  person i.e. assuming that the platoon is stationary.
• 11 years agoHelpfull: Yes(7) No(0)
• Here -----------1F1L
  here is the platoon the distance b/w first and last person is =50 m
  let when last person meet first person platoon move x distance
  so the situation is 2F1f---
  when the platoon move 50 m ahead it means last person is at initial position of
  first person
  As last person run with uniform speed so ratio of speed first and last person remain same while last person moving forward and backward .
  til the time last person handover letter to first person
  first person move=x
  last person move = 50 +x
  when the last person come back
  first person travel in that interval= 50-x
  last travel during backward journey= 50-(50-x)=x
  As both travel all in same time so,
  x/50+x= 50-x/x
  2(x square)= 2500
  x square = 1250
  x= 35.5
  total distance travel by last man =x+50+x=2x+50=70+50=120m
  
• 13 years agoHelpfull: Yes(3) No(6)
• 120.71 metres
• 15 years agoHelpfull: Yes(2) No(2)
• 120 M
• 15 years agoHelpfull: Yes(0) No(3)