Reggie left City X for City W at 08 50. Xylia left City X for City C at the same time. At 11 50, they were 468 km apart. Both maintained a constant speed and reached their destination at the same time at 14 30.
- Find the distance between Cities A and C.
- If Xylia was 57 the speed of Reggie, find Reggie's speed.
(a)
Time taken for Reggie and Xylia to travel 08 50 to 11 50 = 3 h
Time taken for Reggie and Xylia to travel 08 50 to 14 30 = 5 h 40 min = 5
4060 h = 5
23 h
Distance travelled by Reggie and Xylia in 1 hour
= 468 ÷ 3
= 156 km
Total distance
= 5
23 x 156
= 884 km
(b)
Xylia's distance : Reggie's distance
5 : 7
Xylia's speed : Reggie's speed
5 : 7
12 u = 156 km
1 u = 156 ÷ 12 = 13 km
Reggie's speed
= 7 u
= 7 x 13
= 91 km/h
Answer(s): (a) 884 km; (b) 91km/h