A person starts from A to B. After 1hr his tyre gets punctured and
he repairs it for 10 mins and travels the rest of the distant with
30km/hr and arrives 30 min late. had the puncture occured 30 Kms earlier he would been only 15 mins late. Find the ditance betn. A & B and his initial speed.

• Distance = 80kms.
  let the velocity be 'V' kmph, time be 't' hrs and distance between A & B is 'd' km.
  now t = (d/V)
  in first case,
  t + (1/3) = (d1/v) + {(d-d1)/30} assuming after travelling 'd1' tyre got punctured
  second case,
  t+ (1/12) = {(d1+30)/v} + {(d-d1-30)/30} asuming after travelling 'd1+30' tyre got punctured
  now replacing t with (d/v),
  firstcase, (d-d1)/v +(1/3) = (d-d1)/30
  using this in second case we get V = 40 kmph
  now, d1 = v * 1hr = 40km
  using first case,
  (d/40) + (1/3) = 1 + {(d-40)/30)}
  ==> d = 80kms
  
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