Shiyun started jogging from the hypermart to the park at a speed of 96 m/min at 6 a.m. Her sister started jogging from the hypermart later. They were beside each other at 6.45 a.m. and her sister reached the park at 7 a.m. while Shiyun was still 1800 m away. If both of them travelled at a constant speed throughout the journey, what time did her sister leave the hypermart?
1 h = 60 min

From 6 a.m. to 7 a.m. = 1 h
From 6.45 a.m. to 7 a.m. = 15 min

Distance between the hypermart and the park
= 60 x 96 + 1800
= 5760 + 1800
= 7560 m

Distance that Shiyun's sister had to cover to catch up with Shiyun
= 45 x 96
= 4320 m

Distance that Shiyun's sister travelled to the park after catching up with Shiyun
= 7560 - 4320
= 3240 m

Since Shiyun's sister jogged 3240 m in 45 min,
Shiyun's sister average speed
= 3240 ÷ 15
= 216 m/min

Duration that Shiyun's sister took to jog to the park
= 7560 ÷ 216
= 35 min

Time that Shiyun's sister left the hypermart:
35 min before 7 a.m. = 6.25 a.m.

Answer(s): 6.25 a.m.