A software engineer starts from home at 3pm for evening walk. He walks speed of 4kmph on level ground and then at a speed of 3kmph on the uphill and then down the hill at a speed of 6kmph to the level ground and then at a speed of 4kmph to the home
If he reaches home at 9pm. What is the distance on the way

• lets take the level ground be X & uphill be y,
  total time taken 3.0 hr,
  time= dist/speed=x/4+y/3+y/6+x/4=3.0Hr
  3.0=2x/4+3y/6, 3.0=(6x+6y)/12,2(x+y)=12Km will be the answer.
• 14 years agoHelpfull: Yes(19) No(8)
• 24 km
• 14 years agoHelpfull: Yes(8) No(3)
• avg speed=(4+3+6+4)/4=17/4
  time=6
  distance=(17/4)*6=51/2
• 10 years agoHelpfull: Yes(4) No(5)
• due to start point to return the previous point
  time= dist/speed=x/4+y/3+y/6 =6.0Hr
  so, x+y=6*4=24
  if take the level ground be X & uphill be y
• 9 years agoHelpfull: Yes(3) No(0)
• 6 + 6 = 12
• 14 years agoHelpfull: Yes(1) No(1)
• average speed is (4+3+6+4)/4=17/4
  total time is 6
  distace =17/4*6=51/2=25.5 km
• 8 years agoHelpfull: Yes(0) No(1)
• let total distance =x km
  he covers equal distance y km at different speed
  so x/6=4y/(y/4 +y/3+y/6+y/4)
  x=24 km
• 7 years agoHelpfull: Yes(0) No(0)