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On a 750-mi trip from A to B that took a total of 5.5h, that a person took a limousine to theairport, then a plane, and finally a car to reach the

destination. The limousine took as long as the final car trip and the time for connections. The limousine averaged 55 mi/hr, the plane averaged 400 mi/hr and the car averaged 40 mi/hr. The plane traveled four times as far as the limousine and car combined. How long did each part of the trip and the connections take?

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Let c1 and c2 be the times
for connections after Limousine and plane journeys respectively. Let Limousine
take t1 hours to travel S1 miles. Let the plane take t2 hours to travel S2
miles. Then the car takes t3 = (5.5 - t1 - t2 - c1 - c2) hours to cover S3 = (750-
S1 - S2) miles. Let c1+c2 = c = total connection time

S1 =
55 t1 S2 = 400 t2 S3 = 40 t3

t1 + t2 + t3 + c = 5.5 hours, Limousine takes as long as
the car & connections, t1 =
t3 + c -- equation 3 So, t1 + t2 + t1 = 5.5
2 t1 = 5.5 – t2 --- equation 1

The plane travelled distance
4 times as far as limousine and car combined S2 = 4 ( S1 + S3) But S1+S2+S3 = 750 miles,
so S1 + 4 (S1+S3) + S3 = 750 => S1+S3 = 150 miles --- equation 2 S2 = 600 miles t2 = S2/400 mph = 1.5 hours

Substitute t2 in equation
1, we get 2 t1 = 5.5 – 1.5 = 4 =>
t1 = 2 hours So
S1 = 2*55 = 110 miles S3 = 150 – 110 = 40 miles -- using equation 2 t3 = S3/40 mph = 1 hour From equation 3 , we have
t3 + c = 2 hours So C = 2- t3 = 1 hours

I posted this question long back about a week but no one answered this question

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