Cody left City H for City G at 08 47. Xylia left City H for City C at the same time. At 11 47, they were 480 km apart. Both maintained a constant speed and reached their destination at the same time at 14 26.
- Find the distance between Cities A and C.
- If Xylia was 35 the speed of Cody, find Cody's speed.
(a)
Time taken for Cody and Xylia to travel 08 47 to 11 47 = 3 h
Time taken for Cody and Xylia to travel 08 47 to 14 26 = 5 h 39 min = 5
3960 h = 5
1320 h
Distance travelled by Cody and Xylia in 1 hour
= 480 ÷ 3
= 160 km
Total distance
= 5
1320 x 160
= 904 km
(b)
Xylia's distance : Cody's distance
3 : 5
Xylia's speed : Cody's speed
3 : 5
8 u = 160 km
1 u = 160 ÷ 8 = 20 km
Cody's speed
= 5 u
= 5 x 20
= 100 km/h
Answer(s): (a) 904 km; (b) 100km/h