Jaslyn started jogging from the cinema to the park at a speed of 105 m/min at 7 a.m. Her brother started jogging from the cinema later. They were beside each other at 7.36 a.m. and her brother reached the park at 8 a.m. while Jaslyn was still 1800 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 7 a.m. to 8 a.m. = 1 h
From 7.36 a.m. to 8 a.m. = 24 min

Distance between the cinema and the park
= 60 x 105 + 1800
= 6300 + 1800
= 8100 m

Distance that Jaslyn's brother had to cover to catch up with Jaslyn
= 36 x 105
= 3780 m

Distance that Jaslyn's brother travelled to the park after catching up with Jaslyn
= 8100 - 3780
= 4320 m

Since Jaslyn's brother jogged 4320 m in 36 min,
Jaslyn's brother average speed
= 4320 ÷ 24
= 180 m/min

Duration that Jaslyn's brother took to jog to the park
= 8100 ÷ 180
= 45 min

Time that Jaslyn's brother left the cinema:
45 min before 8 a.m. = 7.15 a.m.

Answer(s): 7.15 a.m.