City F and City G is 390 km apart. Peter left City F for City G at 11.00 a.m. travelling at an average speed of 65 km/h. Jack left City F later than Peter and caught up with him at 1.00 p.m. Jack was travelling at a speed of 104 km/h.

1. At what time did Jack leave City F?
2. How much later did Peter arrive in City G than Jack? Express your answer in h as a fraction or mixed number.

(a)
From 11.00 a.m. to 1.00 p.m. = 2 h

Distance that Peter covered when Jack caught up with Peter
= 2 x 65
= 130 km

Time that Jack took to travel 130 km
= 130 ÷ 104
= 1¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
1¹₄ h = 1 h 15 min

Time that Jack left City F:
1 h 15 min before 1.00 p.m. = 11.45 a.m.

(b)
Remaining distance that they had to cover to reach City G
= 390 - 130
= 260 km

Duration that Jack had to travel before reaching City G
= 260 ÷ 104
= 2¹₂ h

Duration that Peter had to travel before reaching City G
= 260 ÷ 65
= 4 h

Duration that Peter took to arrive later than Jack in City G
= 4 - 2¹₂
= 1¹₂ h

Answer(s): (a) 11.45 a.m.; (b) 1¹₂ h