Shiyun started jogging from the market to the park at a speed of 96 m/min at 7 a.m. Her brother started jogging from the market later. They were beside each other at 7.35 a.m. and her brother reached the park at 8 a.m. while Shiyun was still 1600 m away. If both of them travelled at a constant speed throughout the journey, what time did her brother leave the market?
1 h = 60 min

From 7 a.m. to 8 a.m. = 1 h
From 7.35 a.m. to 8 a.m. = 25 min

Distance between the market and the park
= 60 x 96 + 1600
= 5760 + 1600
= 7360 m

Distance that Shiyun's brother had to cover to catch up with Shiyun
= 35 x 96
= 3360 m

Distance that Shiyun's brother travelled to the park after catching up with Shiyun
= 7360 - 3360
= 4000 m

Since Shiyun's brother jogged 4000 m in 35 min,
Shiyun's brother average speed
= 4000 ÷ 25
= 160 m/min

Duration that Shiyun's brother took to jog to the park
= 7360 ÷ 160
= 46 min

Time that Shiyun's brother left the market:
46 min before 8 a.m. = 7.14 a.m.

Answer(s): 7.14 a.m.