Marion started jogging from the cinema to the park at a speed of 105 m/min at 9 a.m. Her brother started jogging from the cinema later. They were beside each other at 9.40 a.m. and her brother reached the park at 10 a.m. while Marion was still 1900 m away. If both of them travelled at a constant speed throughout the journey, what time did her brother leave the cinema?
1 h = 60 min

From 9 a.m. to 10 a.m. = 1 h
From 9.40 a.m. to 10 a.m. = 20 min

Distance between the cinema and the park
= 60 x 105 + 1900
= 6300 + 1900
= 8200 m

Distance that Marion's brother had to cover to catch up with Marion
= 40 x 105
= 4200 m

Distance that Marion's brother travelled to the park after catching up with Marion
= 8200 - 4200
= 4000 m

Since Marion's brother jogged 4000 m in 40 min,
Marion's brother average speed
= 4000 ÷ 20
= 200 m/min

Duration that Marion's brother took to jog to the park
= 8200 ÷ 200
= 41 min

Time that Marion's brother left the cinema:
41 min before 10 a.m. = 9.19 a.m.

Answer(s): 9.19 a.m.