Xuan started jogging from the school to the garden at a speed of 105 m/min at 8 a.m. Her sister started jogging from the school later. They were beside each other at 8.40 a.m. and her sister reached the garden at 9 a.m. while Xuan was still 1400 m away. If both of them travelled at a constant speed throughout the journey, what time did her sister leave the school?
1 h = 60 min

From 8 a.m. to 9 a.m. = 1 h
From 8.40 a.m. to 9 a.m. = 20 min

Distance between the school and the garden
= 60 x 105 + 1400
= 6300 + 1400
= 7700 m

Distance that Xuan's sister had to cover to catch up with Xuan
= 40 x 105
= 4200 m

Distance that Xuan's sister travelled to the garden after catching up with Xuan
= 7700 - 4200
= 3500 m

Since Xuan's sister jogged 3500 m in 40 min,
Xuan's sister average speed
= 3500 ÷ 20
= 175 m/min

Duration that Xuan's sister took to jog to the garden
= 7700 ÷ 175
= 44 min

Time that Xuan's sister left the school:
44 min before 9 a.m. = 8.16 a.m.

Answer(s): 8.16 a.m.