Betty started jogging from the park to the garden at a speed of 96 m/min at 5 a.m. Her brother started jogging from the park later. They were beside each other at 5.45 a.m. and her brother reached the garden at 6 a.m. while Betty was still 1800 m away. If both of them travelled at a constant speed throughout the journey, what time did her brother leave the park?
1 h = 60 min

From 5 a.m. to 6 a.m. = 1 h
From 5.45 a.m. to 6 a.m. = 15 min

Distance between the park and the garden
= 60 x 96 + 1800
= 5760 + 1800
= 7560 m

Distance that Betty's brother had to cover to catch up with Betty
= 45 x 96
= 4320 m

Distance that Betty's brother travelled to the garden after catching up with Betty
= 7560 - 4320
= 3240 m

Since Betty's brother jogged 3240 m in 45 min,
Betty's brother average speed
= 3240 ÷ 15
= 216 m/min

Duration that Betty's brother took to jog to the garden
= 7560 ÷ 216
= 35 min

Time that Betty's brother left the park:
35 min before 6 a.m. = 5.25 a.m.

Answer(s): 5.25 a.m.