A software engineer starts from home at 3pm for every walk on a route which has level road for some distance and a then a hillock.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 of6kmph to the level ground and then at a speed of 4kmph to the home if he reaches home at 9pm.what is the distance one way of his route?
(A)12km (B)15km (C)18km (D)24km(E)Data Inadequate

• time=distance/speed
  so,
  let the distance to the level road from home be x km
  and the distance to the hillock from the level road be y km
  so, (x/4+y/3)+(y/6+x/4)=6
  x+y=12 km.
• 13 years agoHelpfull: Yes(37) No(4)
• t1 be time for level road
  ,t2 for upward and t3 for downward as given t2=2t3 and t1+t2+t1+t3=6
  we get, 4t1+3t2=12 ans
• 13 years agoHelpfull: Yes(10) No(0)
• answer will be 24 km
  apply the formula for average speed 2ab/a+b for all the 4 speeds
  2x4x3/(4+3)=24/7
  2x6x4/10=24/5
  so at last again apply the formula for average speed
  speed=4 kmph
  so distance=speed x time
  =4x6=24kmph
• 13 years agoHelpfull: Yes(6) No(23)
• 1/2(5/12+4/12)=3/8 ans=3/8
• 13 years agoHelpfull: Yes(2) No(16)