Question

At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist heading north is riding 4 km/hour faster than the bicyclist heading south. At 10:30, they are 39 km apart. Find the two bicyclists’ rates.
(A)northbound bicyclist = 15 km/h; southbound bicyclist = 11 km/h
(B)northbound bicyclist = 17 km/h; southbound bicyclist = 11 km/h
(C)northbound bicyclist = 15 km/h; southbound bicyclist = 10 km/h
(D)northbound bicyclist = 16 km/h; southbound bicyclist = 12 km/h

Answer 1

Let x be the speed of the cyclist riding south.

Speed of the cyclist reading north = x + 4.

9:00 - 10:30 = 1.5 hours has passed.

As Distance = Speed x Time
Therefore,
Distance travelled by cyclist riding south = (x+4)(1.5) = 1.5x + 6
Distance travelled by cyclist riding north = (x)(1.5) + 1.5x

Total distance between them at 10:30 = Distance travelled by cyclist riding south + Distance travelled by cyclist riding north
= 1.5x + 6 + 1.5x = 3x + 6

3x + 6 = 39
x = 11 = Southbound Cyclist

Northbound Cyclist = 11+4 = 15.

Answer: A

PS: Always helps to use algebra