Latest Infosys Aptitude Question SOLUTION: 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 sche

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(#M40006805) INFOSYS Numerical Ability question Keep an EYE

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?

Asked In Infosys jyothi (12 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

praveen  (12 years ago)

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.

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Amalesh Jana  (11 years ago)

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

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naman  (5 years ago)

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

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