A man starts walking at 3 pm, He walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill, 6 km/hr downhill and then 4 km/hr on level ground to reach home at 9 pm. What is the distance covered on one way?

• if x km is level ground and y km is hill road, then
  x/4 + y/3 +y/6+x/4= 6
  6(x+y)=72
  x+y=12 km
  
  so one side distance is 12 kms.
• 12 years agoHelpfull: Yes(84) No(24)
• as the man contineously with diff speed
  find avg of speed by
  no. of speed/(1/s1+1/s2+1/s3+1/s4)=avg speed
  4/(1/4+1/3+1/6+1/4)=4km/hr
  time=9-3=6
  6*4=24kms
• 12 years agoHelpfull: Yes(27) No(7)
• dear garima,you have taken x twice in the equation and also y....so keeping that frame in mind your total distance in one side should be 2(x+y)= 24 kms
• 12 years agoHelpfull: Yes(17) No(19)
• incomplete ques.as distance on plane is not given to be same
• 10 years agoHelpfull: Yes(3) No(0)
• here level ground and hill road
  
  let
  level ground=x
  hill road =y
  starting time=3 p.m
  ending time=9 p.m
  therefore time=9-3
  =6
  let eq
  x/4+y/3+y/6+x/4=6
  (6x+6y)/12=6
  6x+6y=70
  6(x+y)=70
  x+y=70/6
  +y=12kms
• 10 years agoHelpfull: Yes(2) No(0)
• The question is " A man starts walking ***from home*** at 3 pm . ha walks at a speed of 4 km/hr on level ground and at a speed of 3 km/hr on uphill , 6 km/hr downhill and then 4 km/hr on level ground to reach ***back*** home at 9 pm. What is the distance covered on one way?"
  
  Now the ans is 12 as GARIMA explained...
  the question was not properly explained
• 6 years agoHelpfull: Yes(1) No(0)