At 7 a.m., Howard started from City S and travelled towards City T and his speed remained constant throughout. At 9 a.m., Tina started her journey from City S towards City T at an average speed of 84 km/h. Tina overtook Howard at 1 p.m. After overtaking, Tina carried on her journey at the same average speed and reached at City T at 1:50 p.m.

1. Find Howard'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 Howard
= 4 x 84
= 336 km

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

Howard's average speed
= 336 ÷ 6
= 56 km/h

(b)
From 9 a.m. to 1:50 p.m. = 4 h 50 min

1 h = 60 min
4 h 50 min = 4 ⁵⁰ ₆₀
 h = 4 ⁵₆ h

Distance between the two cities
= 4 ⁵₆ x 84
= ²⁹₆ x 84
= 406 km

Answer(s): (a) 56 km/h; (b) 406 km