Jade started jogging from the cafe to the park at a speed of 105 m/min at 6 a.m. Her brother started jogging from the cafe later. They were beside each other at 6.36 a.m. and her brother reached the park at 7 a.m. while Jade was still 1800 m away. If both of them travelled at a constant speed throughout the journey, what time did her brother leave the cafe?
1 h = 60 min

From 6 a.m. to 7 a.m. = 1 h
From 6.36 a.m. to 7 a.m. = 24 min

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

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

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

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

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

Time that Jade's brother left the cafe:
45 min before 7 a.m. = 6.15 a.m.

Answer(s): 6.15 a.m.