A software engineer starts from home at 3 PM for evening 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 of 6 kmph to the level ground and then at a speed of 4kmph to the home If he reaches home at 9 PM. What is the distance one way of his route?
option
(A) 12 Km
(B) 15 Km
(C) 18 Km
(D) 24 Km
(E) Data Inadequate

• ans is 12kms....
  let the distance travelled in level ground is x kms in t hrs...
  then x=4t....
  let distance travelled in the hillock is y kms in (t+a)hrs....
  then y=3(t+a)....
  back to the house distanc in hillock is 6(t-a)kms...
  in the level ground is 4t....
  total time taken to reach home is 6hrs...
  t+(t+a)+(t-a)+t=6 there4 t=3/2....
  distance from and is same there4 by equating 4t+3(t+a)=4t+6(t-a)
  ie...a=1/2...
  distance from house to hillock is 12kms....
• 11 years agoHelpfull: Yes(28) No(6)
• ans is 12. sorry..
• 11 years agoHelpfull: Yes(2) No(1)
• i think ans should be 10km.. but its not in the option :(
• 11 years agoHelpfull: Yes(0) No(5)
• @ROJA
  your method is right
  answer is wrong
  Answer is 24
  
  X=4T---------2 TIMES -----> GIVES 12( 6 EACH )
  Y=6(T-A) ------> GIVES 6
  Y=3(T+A)-------> GIVES 6
  therefore total 24
  (value of A = 1/2 && T = 3/2)
• 6 years agoHelpfull: Yes(0) No(0)