A starts 3mins after B for a place 4.5km distant B, on reaching his destination, immediately returns and after walking a km meets A.A can walk 1km in 18 mins.what is the speed of B?

• The answer is 5kmph. This Question has full of walking. A traveled for 3.5 kilometers in 66 minutes (3.5X18=63=>because of the delay=>63+3=66)but B traveled 5.5 km therefore 66/5.5=12mins. i.e. B takes 12 mins to walk a km. so his walking speed is 60/12 = 5kmph
  
• 12 years agoHelpfull: Yes(5) No(0)
• let b travel t time
  t=(4.5+1)/x.........(1)
  for a
  t-3=(4.5-1)/1.....(2)
  on solving
  3=(5.5/x)-3.5
  3x=5.5-3.5x
  6.5x=5.5
  x=5.5/6.5=55/65 m/min ans
  
• 12 years agoHelpfull: Yes(3) No(3)
• ans is 5km/hr..
  
• 12 years agoHelpfull: Yes(2) No(0)
• i think the pattern has again changed for month of oct. so frnds pls help ,solve the ques and upload the answers.
• 12 years agoHelpfull: Yes(0) No(2)
• answer is 4.85 km
  
• 12 years agoHelpfull: Yes(0) No(3)
• thank u so much admin....pls solve each and every prob.....its urgent....
• 12 years agoHelpfull: Yes(0) No(0)