On the first part of a cyclists journey, the cyclist's speed was constant at 22km/h. On the second part of the journey, the cyclist's speed was constant at 33km/h. The total average speed of the cyclist for both parts was 25km/h. What was the total distance of both parts, if the second part was 38.5km shorter than the first part.

answered Jan 5 by Level 10 User (55,380 points)

Let the distances for the two parts of the journey be F and S, and S=F-38.5.

Let T1 be the time for the first part of the journey and T2 the time for the second part.

F=22T1, S=33T2=F-38.5; F+S=25(T1+T2); 2F-38.5=25(F/22+(F-38.5)/33).

Multiply the last equation by 66: 132F-2541=75F+50F-1925.

(132-75-50)F=2541-1925; 7F=616 and F=88km, so S=F-38.5=49.5km.

Length of 1st part of journey=88km and 2nd part of journey=49.5km.

CHECK

T1=88/22=4hr; T2=49.5/33=1.5hr. T1+T2=5.5hr; total distance=137.5km.

Average speed=137.5/5.5=25km/hr. So average speed checks out.

answered Jan 5 by Top Rated User (424,700 points)