A man is going to a party.He travells for 2hrs he gets a puncture.Changing tyres takes 10min.the rest of the journey he travells at 30 miles per hr.he reaches 30min behind the schedule.He thinks to himself that the puncture had occur 30miles later,he would have been only 15min late.find the total distance travelled by him?

• 3hrs 25 mins ,let the total time be x,then
  as he thinks before 30miles before,then the total time he travelled by 15mins late is 2hr +10mins+1hr - 15mins =2hr55mins,bt he was late by 30 mins so the total time he travels in will be 3hr 25mins.
• 12 years agoHelpfull: Yes(0) No(4)
• distance betn spot of puncture and event is 37.5 miles.
  60.....30
  15...15*30/60=7.5
  
  so the distance is 30+7.5=37.5 miles
• 11 years agoHelpfull: Yes(0) No(0)
• Let normal speed before puncture be v and distance left to cover is d...If there was no puncture then total time will be 2 + d/v....Now the actual case total time is 2 + 10min+ d/30
  Equating 2+d/v +30 = 2+10 + d/30 => d/30 - d/v = 1/3hr
  Supposed case time is 2+30/v + 10+ (d-30)/30
  this time +15 mins will be equal to 2nd case
  So 2+30/v + 10+ (d-30)/30 +15 = 2+10+d/30
  =>30/v = 1- 1/4 => v = 40miles/hr
  Substituting :: d/30 - d/40 = 1/3
  =>d =40miles
  Total distance = 40*2 + 40 = 120 miles
• 6 years agoHelpfull: Yes(0) No(0)