Jean started jogging from the playground to the park at a speed of 105 m/min at 5 a.m. Her sister started jogging from the playground later. They were beside each other at 5.40 a.m. and her sister reached the park at 6 a.m. while Jean was still 1900 m away. If both of them travelled at a constant speed throughout the journey, what time did her sister leave the playground?
1 h = 60 min

From 5 a.m. to 6 a.m. = 1 h
From 5.40 a.m. to 6 a.m. = 20 min

Distance between the playground and the park
= 60 x 105 + 1900
= 6300 + 1900
= 8200 m

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

Distance that Jean's sister travelled to the park after catching up with Jean
= 8200 - 4200
= 4000 m

Since Jean's sister jogged 4000 m in 40 min,
Jean's sister average speed
= 4000 ÷ 20
= 200 m/min

Duration that Jean's sister took to jog to the park
= 8200 ÷ 200
= 41 min

Time that Jean's sister left the playground:
41 min before 6 a.m. = 5.19 a.m.

Answer(s): 5.19 a.m.