A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance T between the two cities?  Answers.
A) 48  B) 72  C) 96  D) 120  E) 192
Hmmm I did this question a while ago and got it wrong so trying again   Rate x time = distance.   For the initial trip lets the distance T to the midpoint be represented by P.   EQN1: R x 2 = P   For the second trip we know one bus left late and one left early. Together this is just a tricky way of saying one bus left an hour after the other. We know the total trip takes 4 hours (since getting to P is 2 hours). The second trip can be represented by:   Since the trip takes 4 hours if a bus leaves one hour early, the reminaining 3 hours are split between the two buses, ie. 1 + 3/2 = 2.5   EQN2: R x 2.5 = P + 24   EQN2-EQN1 : 0.5R=24   R=48   So the distance is rate x time = 48 x 4 = 192.   ANS = 192
<result>E</result>
A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance T between the two cities?