Q. A software engineer returns from America. As he is fat he decided to have evening walk.....he started at 3pm. he walks along the road at 4km/hr for some time then he climbs a upward slope area at 3km/hr then downwards at the rate of 6km/hr. then back to the home through the road at 4km/hr. what is the distance he covered in one way if he reaches back home at 9pm.

• Let the dist traveled on level ground = x and on the hill be y.
  x/4 + y/3 + y/6 + x/4 = 6
  (x+y)/2 = 6km
  
• 9 years agoHelpfull: Yes(4) No(2)
• ans is 6 km
• 10 years agoHelpfull: Yes(3) No(4)
• 12 km (one way )( total distance:12* 2 = 24 km)
• 10 years agoHelpfull: Yes(2) No(13)
• here we dont know exact dis....so ..let consider the dis travelled on road is X ...and dis on upward slope area is Y...so....toatal time is. 6 hrs...then...X/4+ Y/3 + X/4 +Y/6 = 6
  (X+Y)/2 = 6...here we need distance he covered in one way.....so X+Y = 12.....
• 9 years agoHelpfull: Yes(2) No(0)
• his speeds are 4km/hr , 3km/hr ,6km/hr , again back to home with a speed of 4km/hr.
  as he is asking only one way distance,we need not consider the return speed 4km/hr to home.
  his journey is totally 3pm to 9pm ------> 6hrs
  hence if we take 4km/hr * 2hrs + (3km/hr*1hr) + 6km/hr * 1hr + (4km/hr *2 hrs) = 17 km of distance
  neglecting the return distance of (4km/hr *2 hrs) ------> 17km - 8km = 9km in one way
• 9 years agoHelpfull: Yes(1) No(4)
• Let the distance travelled by him is: x
  Total time taken to walk : 6 hrs
  x/4+x/3+x/6+x/4=6
  6x+8x+4x+6x=144
  then x=6km
  ans is 6km
• 9 years agoHelpfull: Yes(1) No(0)