At 7 a.m., Oscar started from City F and travelled towards City G and his speed remained constant throughout. At 9 a.m., Tina started her journey from City F towards City G at an average speed of 66 km/h. Tina overtook Oscar at 1 p.m. After overtaking, Tina carried on her journey at the same average speed and reached at City G at 4:50 p.m.

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

(a)
From 9 a.m. to 1 p.m. = 4 h

Distance that Tina had travelled until Tina overtook Oscar
= 4 x 66
= 264 km

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

Oscar's average speed
= 264 ÷ 6
= 44 km/h

(b)
From 9 a.m. to 4:50 p.m. = 7 h 50 min

1 h = 60 min
7 h 50 min = 7 ⁵⁰ ₆₀
 h = 7 ⁵₆ h

Distance between the two cities
= 7 ⁵₆ x 66
= ⁴⁷₆ x 66
= 517 km

Answer(s): (a) 44 km/h; (b) 517 km