At 7 a.m., Jack started from City R and travelled towards City S and his speed remained constant throughout. At 10 a.m., Xylia started her journey from City R towards City S at an average speed of 72 km/h. Xylia overtook Jack at 1 p.m. After overtaking, Xylia carried on her journey at the same average speed and reached at City S at 6:15 p.m.

1. Find Jack's average speed in km/h.
2. What is the distance between the two cities?

(a)
From 10 a.m. to 1 p.m. = 3 h

Distance that Xylia had travelled until Xylia overtook Jack
= 3 x 72
= 216 km

From 7 a.m. to 1 p.m. = 6 h

Jack's average speed
= 216 ÷ 6
= 36 km/h

(b)
From 10 a.m. to 6:15 p.m. = 8 h 15 min

1 h = 60 min
8 h 15 min = 8 ¹⁵ ₆₀
 h = 8 ¹₄ h

Distance between the two cities
= 8 ¹₄ x 72
= ³³₄ x 72
= 594 km

Answer(s): (a) 36 km/h; (b) 594 km