Latest Infosys Aptitude Question SOLUTION: A man is going to a wedding party. He travels for 2hrs when he gets a puncture. Changing tyres takes 10mins. The rest of the journey he travels at 30 miles/hr. He reaches 30mins b

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A man is going to a wedding party. He travels for 2hrs when he gets a puncture. Changing tyres takes 10mins. The rest of the journey he travels at 30 miles/hr. He reaches 30mins behind schedule. He thinks to himself that if the puncture had occurred 30miles later, he would have been only 15mins late. Find the total distance traveled by the man

Asked In Infosys Aditi Sharma (12 years ago)

Unsolved Read Solution (2)

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Read Solution (2) :

Sulekha  (12 years ago)

120 miles.
normal speed 40 mph.
Normal time = 3 hrs
Distance travelled in 2 hrs = 80 miles.
Time taken for last 40 miles = 40/30 =4/3 hrs .. 1 hr 20 mis.
Time taken to replace tyre = 10 mins.

so late by 30 mins.

If puncture would have occurred after 110 miles.
Time taken for 110 miles = 110/40 = 2 hrs 45 mins.
Time taken for balance 10 miles @ 30 mph= 20 mis.
Time for changing tyre = 10 mins.

so would have been late by 15 mins.

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nagini  (8 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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