A bus travels at a constant speed from point A to point B, which are 600 km apart. Some time later, a motorcyclist begins the trip from A to B, traveling at 100 km/h. The motorcyclist overtakes the bus at 150 km from A, and continues on to point B. The motorcyclist arrives at B, rests for 1 hour, and begins the return trip to A, crossing paths with the bus. When the bus arrives at B, the motorcyclist is 200 km from B. How far from B did the motorcycle and bus meet the second time?

Step-by-step explanation:

the motorcyclist starts x hours after the bus.

it takes the motorcyclist 1.5 hours to meet the bus the first time, 6 hours to reach B, rest 1 hour and go another 2 hours before the bus arrives at B.

so, the bus needs

x + 9 hours to reach B.

x + 1.5 hours for 150 km (for the first meeting with the motorcyclist).

the speed of the bus is 600/(x + 9) km/h or 150/(x + 1.5) km/h.

so,

600/(x + 9) = 150/(x + 1.5)

4/(x + 9) = 1/(x + 1.5)

4(x + 1.5)/(x + 9) = 1

4(x + 1.5) = x + 9

4x + 6 = x + 9

3x + 6 = 9

3x = 3

x = 1

the motorcyclist starts 1 hour after the bus.

the speed of the bus is

600/(x + 9) = 600/(1 + 9) = 600/10 = 60 km/h

when the motorcyclist is starting its return trip, he is 7 hours after his start, and the bus is 7+x = 7+1 = 8 hours after its start, that means at

8×60 = 480 km of its total trip of 600 km.

120 km are still left for the bus to go.

from that point on both travel the same time y going with their speeds until they meet the second time.

the motorcyclist travels 100y km building up distance from B, while the bus travels 60y km reducing the 120 km distance from B.

when they meet the second time, the distance from B for the motorcyclist is the same as the remaining distance for the bus.

so,

100y = 120 - 60y

160y = 120

y = 120/160 = 3/4 = 0.75 hours

so, they meet at

100 × 0.75 = 75 km from B.

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