City Z and City A is 434 km apart. Gabriel left City Z for City A at 11.00 a.m. travelling at an average speed of 62 km/h. Peter left City Z later than Gabriel and caught up with him at 1.00 p.m. Peter was travelling at a speed of 93 km/h.

1. At what time did Peter leave City Z?
2. How much later did Gabriel arrive in City A than Peter? 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 Gabriel covered when Peter caught up with Gabriel
= 2 x 62
= 124 km

Time that Peter took to travel 124 km
= 124 ÷ 93
= 1¹₃ h

1 h = 60 min
¹₃ h = ¹₃ x 60 = 20 min
1¹₃ h = 1 h 20 min

Time that Peter left City Z:
1 h 20 min before 1.00 p.m. = 11.40 a.m.

(b)
Remaining distance that they had to cover to reach City A
= 434 - 124
= 310 km

Duration that Peter had to travel before reaching City A
= 310 ÷ 93
= 3¹₃ h

Duration that Gabriel had to travel before reaching City A
= 310 ÷ 62
= 5 h

Duration that Gabriel took to arrive later than Peter in City A
= 5 - 3¹₃
= 1²₃ h

Answer(s): (a) 11.40 a.m.; (b) 1²₃ h