At 9 a.m., Owen started from City S and travelled towards City T and his speed remained constant throughout. At 10 a.m., Xuan started her journey from City S towards City T at an average speed of 90 km/h. Xuan overtook Owen at 2 p.m. After overtaking, Xuan carried on her journey at the same average speed and reached at City T at 4:50 p.m.

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

(a)
From 10 a.m. to 2 p.m. = 4 h

Distance that Xuan had travelled until Xuan overtook Owen
= 4 x 90
= 360 km

From 9 a.m. to 2 p.m. = 5 h

Owen's average speed
= 360 ÷ 5
= 72 km/h

(b)
From 10 a.m. to 4:50 p.m. = 6 h 50 min

1 h = 60 min
6 h 50 min = 6 ⁵⁰ ₆₀
 h = 6 ⁵₆ h

Distance between the two cities
= 6 ⁵₆ x 90
= ⁴¹₆ x 90
= 615 km

Answer(s): (a) 72 km/h; (b) 615 km