Yen started jogging from the hypermart to the garden at a speed of 108 m/min at 8 a.m. Her sister started jogging from the hypermart later. They were beside each other at 8.35 a.m. and her sister reached the garden at 9 a.m. while Yen 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 8 a.m. to 9 a.m. = 1 h
From 8.35 a.m. to 9 a.m. = 25 min

Distance between the hypermart and the garden
= 60 x 108 + 1800
= 6480 + 1800
= 8280 m

Distance that Yen's sister had to cover to catch up with Yen
= 35 x 108
= 3780 m

Distance that Yen's sister travelled to the garden after catching up with Yen
= 8280 - 3780
= 4500 m

Since Yen's sister jogged 4500 m in 35 min,
Yen's sister average speed
= 4500 ÷ 25
= 180 m/min

Duration that Yen's sister took to jog to the garden
= 8280 ÷ 180
= 46 min

Time that Yen's sister left the hypermart:
46 min before 9 a.m. = 8.14 a.m.

Answer(s): 8.14 a.m.