Lynn started jogging from the hypermart to the park at a speed of 105 m/min at 5 a.m. Her cousin started jogging from the hypermart later. They were beside each other at 5.41 a.m. and her cousin reached the park at 6 a.m. while Lynn was still 1900 m away. If both of them travelled at a constant speed throughout the journey, what time did her cousin leave the hypermart?
1 h = 60 min

From 5 a.m. to 6 a.m. = 1 h
From 5.41 a.m. to 6 a.m. = 19 min

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

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

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

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

Duration that Lynn's cousin took to jog to the park
= 8200 ÷ 205
= 40 min

Time that Lynn's cousin left the hypermart:
40 min before 6 a.m. = 5.20 a.m.

Answer(s): 5.20 a.m.