Question

two cyclists, 112 miles apart riding toward each other at the same time. One cycles 3 times as fast as the other. if they meet 4 hours later, what is the speed (in mi/h) of the faster cyclist?

Answer 1

distance= 112 miles apart

time = 4 hours

Speed of a = 3 *speed of b

Distance covered by a = 3 * distance covered by b

Speed = distance / time

3b+b = 112

4b = 112

b= 112/4

b= 28 miles

a = 3b

a = 3 *28

a= 84 miles

Since both travelled for 4 hours:

Speed of A = 84 /4 = 21 mph

Speed of b= 28 /4 = 7 mph

Speed of the faster cyclist : 21 mph