City R and City S is 544 km apart. Tim left City R for City S at 10.00 a.m. travelling at an average speed of 68 km/h. Zeph left City R later than Tim and caught up with him at 2.00 p.m. Zeph was travelling at a speed of 102 km/h.

1. At what time did Zeph leave City R?
2. How much later did Tim arrive in City S than Zeph? Express your answer in h as a fraction or mixed number.

(a)
From 10.00 a.m. to 2.00 p.m. = 4 h

Distance that Tim covered when Zeph caught up with Tim
= 4 x 68
= 272 km

Time that Zeph took to travel 272 km
= 272 ÷ 102
= 2²₃ h

1 h = 60 min
²₃ h = ²₃ x 60 = 40 min
2²₃ h = 2 h 40 min

Time that Zeph left City R:
2 h 40 min before 2.00 p.m. = 11.20 a.m.

(b)
Remaining distance that they had to cover to reach City S
= 544 - 272
= 272 km

Duration that Zeph had to travel before reaching City S
= 272 ÷ 102
= 2²₃ h

Duration that Tim had to travel before reaching City S
= 272 ÷ 68
= 4 h

Duration that Tim took to arrive later than Zeph in City S
= 4 - 2²₃
= 1¹₃ h

Answer(s): (a) 11.20 a.m.; (b) 1¹₃ h