A person started his journey in the morning. At 11 A.M, he covered 3/8 th of his journey and on the same day at 4 : 30 P.M, he covered 5/6 th of his journey. At what time did he start his journey?

• Let total journey = x km
  Time between 11 AM and 4:30 PM = 5 hours 30 min = 11/2 hours
  Distance travelled between 11 AM and 4:30 PM = 5x/6 – 3x/8 = 11x/24
  Speed of Person = (11x/24)/(11/2) = x/12 km/hr
  Time required for covering 3/8th of the journey = (3x/8)/(x/12) = 9/2 hours
  Hence he started the journey at 11 – 9/2 = 6:30 AM
• 7 years agoHelpfull: Yes(11) No(0)
• Let total journey = 120 km(let)
  Time between 11 AM and 4:30 PM = 5 hours 30 min = 11/2 hours
  Distance travelled till 11 AM =3/8*120 = 45km
  Distance travelled till 4:30 PM =5/6*120 = 100km
  so , distance travelled in 5.5 hrs = 55km => 10 km/hr
  
  hence, till 11am he travelled 45 km i.e. 45/10 = 4.5 hrs berfore 11am = 6:30 AM
• 5 years agoHelpfull: Yes(1) No(0)
• since his speed is same trough out the journey
  let spped =v
  let total distance is t
  (3/8t)/v=11-t.....................1
  (5/6t)/v=9/2+1+11-t ................2
  
  solve and get the answer
• 7 years agoHelpfull: Yes(0) No(1)