Question

On a 750-mi trip from A to B that took a total of 5.5h, that a person took a limousine to the
airport, 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?

Answer 1

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