a,b,c walk 1km in 5 min,8min and 10min respectively.c starts walking from a point,at a certain time,b starts from the same point 1 min later and a starts from the same point 2 min later than c.then a meets b and c after.?

• A walks 1 km in 5 min.
  Then, distance covered by A in 1 min = 1000/5 = 200 m
  
  B walks 1 km in 8 min.
  Then, distance covered by B in 1 min = 1000/8 = 125 m
  
  C walks 1 km in 10 min.
  Then, distance covered by C in 1 min = 1000/101 = 100 m
  
  Let A meets B and C in x and y min, respectively.
  
  Then
  Distance covered by C in(x + 2) min = Distance covered by A in x min
  100(x + 2) = 200 x
  100 x + 200 = 200 x
  200 = 100 x
  x = 200/100 = 2 min
  
  for A and B
  Distance covered by B in (y + 1) min = Distance covered by A in y min
  125(y + 1) = 200y
  125y + 125 = 200y
  125 = 200y - 125y
  125 = 75y
  y =5/3 minut.
• 7 years agoHelpfull: Yes(6) No(0)
• speed of a,b,c =1/5,1/8,1/10 km/min respectively
  c covers in 2mins=2/10 km
  b covers in 1 min=1/8km
  then a started
  speed diff between a & b=(1/5-1/8)=3/40km/min
  in t=d/s=(1/8) / (3/40) =5/3 min =1min 40 sec a will cover the initial gap between a&b (ans)
  speed diff between a & c=1/10 km/min
  in t=d/s=(1/5) / (1/10) =2 min a will cover the initial gap between a&c (ans)
• 7 years agoHelpfull: Yes(1) No(0)
• speed of a=1/3, b =5/24 ,c=1/6 m/s [ 1km=100m and 1 min=60s]
  consider the time as
  for c= t+120 [2 min =120s ]
  for b=t+60 [ 1min=60s ]
  for a=t
  a meets b , Then both distance are equal.so we can use speed*time.
  [t+60]*5/24=t*24
  t=100s [1min 40s]
  a meets c
  [t+120]*1/6=t*1/3
  t=120s [2min].
• 7 years agoHelpfull: Yes(0) No(0)