Betty started jogging from the hypermart to the park at a speed of 115 m/min at 9 a.m. Her cousin started jogging from the hypermart later. They were beside each other at 9.43 a.m. and her cousin reached the park at 10 a.m. while Betty was still 1700 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 9 a.m. to 10 a.m. = 1 h
From 9.43 a.m. to 10 a.m. = 17 min

Distance between the hypermart and the park
= 60 x 115 + 1700
= 6900 + 1700
= 8600 m

Distance that Betty's cousin had to cover to catch up with Betty
= 43 x 115
= 4945 m

Distance that Betty's cousin travelled to the park after catching up with Betty
= 8600 - 4945
= 3655 m

Since Betty's cousin jogged 3655 m in 43 min,
Betty's cousin average speed
= 3655 ÷ 17
= 215 m/min

Duration that Betty's cousin took to jog to the park
= 8600 ÷ 215
= 40 min

Time that Betty's cousin left the hypermart:
40 min before 10 a.m. = 9.20 a.m.

Answer(s): 9.20 a.m.