At 12 50, Justin left Town S for Town T, driving at a speed of 76 km/h. At 15 50, Albert also left Town S for Town T driving at a certain speed. Both of them did not change their speed throughout the journey. At 18 50, both of them passed a park that was 228 km away from Town T. How many minutes earlier did Albert reach Town T than Justin?

Distance that Justin travelled
= 6 x 76
= 456 km

Distance between Town S and Town T
= 456 + 228
= 684 km

Since Albert passed the park at the same time as Justin, it means that Albert travelled 456 km in 3 hours.

Albert's speed during the journey
= 456 ÷ 3
= 152 km/h

Duration that Albert took to reach Town T after passing the park
= 228 ÷ 152
= 1.5 h

Duration that Justin took to reach Town T after passing the park
= 228 ÷ 76
= 3 h

Difference between the time they took to reach Town T
= 3 - 1.5
= 1.5 h

1 h = 60 min
1.5 h = 1.5 x 60 = 90 min

Albert reached Town T 90 minutes earlier than Justin.

Answer(s): 90 min