Linda started jogging from the playground to the park at a speed of 105 m/min at 6 a.m. Her sister started jogging from the playground later. They were beside each other at 6.41 a.m. and her sister reached the park at 7 a.m. while Linda 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 6 a.m. to 7 a.m. = 1 h
From 6.41 a.m. to 7 a.m. = 19 min

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

Distance that Linda's sister had to cover to catch up with Linda
= 41 x 105
= 4305 m

Distance that Linda's sister travelled to the park after catching up with Linda
= 8200 - 4305
= 3895 m

Since Linda's sister jogged 3895 m in 41 min,
Linda's sister average speed
= 3895 ÷ 19
= 205 m/min

Duration that Linda's sister took to jog to the park
= 8200 ÷ 205
= 40 min

Time that Linda's sister left the playground:
40 min before 7 a.m. = 6.20 a.m.

Answer(s): 6.20 a.m.