# Wedding party Distance Question

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

#### gaurav.pruthi88 Profile Answers by gaurav.pruthi88

• Nov 14th, 2013

82.5miles

#### sachin

• Nov 27th, 2013

112.5 miles

#### Visa

• Dec 14th, 2013

90 miles

#### Visa Ram Profile Answers by Visa Ram

• Dec 14th, 2013

90 miles

#### RB

• Jan 4th, 2014

97.5 miles

#### Jitender

• Jan 10th, 2014

120 miles.

#### Olga Sermon

• Sep 8th, 2014

We can build the equations on the path. We will call his original speed S, the time he spent traveling after the puncture T1, and the hypothetical time he would have spent traveling if the puncture occurred 30 mins later T2.

So, first his actual path:

2*S+T1*30=(his ideal path, as if no puncture occurred:)=S*(2+1/6+T1-1/2)=
(his hypothetical path, with puncture 20 mins later)=S*(2.5+1/6+T2-1/4)

so, we have S*(2+1/6+T1-1/2)=S*(2.5+1/6+T2-1/4) ----> T1=T2+3/4
from this 2*S+T1*30=S*(2.5+1/6+T2-1/4) ---> 2*S+T1*30=S*(T1+3/4)
----->S=45
Now we out it into our first equation:
90+30*T1=45(1+1/2+1/6+T1)
We now know that T1=1.22, and the path is 91.22

#### AusskieMike

• Sep 10th, 2014

Not enough information. If we assume he drove at a constant velocity for the first 2 hour leg, we can calculate what it is because the difference between the time it takes to travel 30 miles at his first-leg velocity is 15 minutes quicker than the time it takes to travel 30 miles at his second-leg velocity.

15 min = (30 miles/30mph) - (30 miles/ first-leg-velocity)
first-leg-velocity = 40 mph

- 2 hours at this speed would make the first leg of the trip 80 miles

- the second leg of the trip is greater than 30 miles because of the statement about the puncture possibly occurring 30 miles later, which would make the total trip greater than 110 miles

- we have no information about the time it took to travel the second leg, and no further information about the distance

- with only a velocity and a idea that it is greater than 30 miles, we cannot find the exact distance of the entire trip
Distance > 110 miles

• Jan 31st, 2015

180miles

#### abhishek

• Feb 14th, 2015

266.66

#### parth

• Aug 18th, 2015

102.5

#### premchand

• Sep 10th, 2015

120 m

#### devika

• Sep 15th, 2015

36000m

• Oct 28th, 2015

35 miles

#### Zooey Profile Answers by Zooey

• Dec 10th, 2015

Since the original speed is 15mins faster than the second speed for 30miles, it means it takes 45mins for 30miles -> 40miles per hr ->80miles on the first part of the journey.

With 30mins behind schedule with 10mins changing the tyres, he is spending 20mins extra with second speed than first speed - for each hour, its making 15 mins difference, it takes 20mins/15mins hour for the second part of the journey -> 20mins/15mins *30miles =40miles

Total distance =120miles

#### Zooey Profile Answers by Zooey

• Dec 10th, 2015

120miles

#### FSS

• Nov 4th, 2016

410 miles.

#### nave een

• Dec 30th, 2016

30min 30miles
60min 60miles
90min 90miles
120min 120miles

Total 120 miles