City S and City T is 539 km apart. Ivan left City S for City T at 9.00 a.m. travelling at an average speed of 77 km/h. Ryan left City S later than Ivan and caught up with him at 1.00 p.m. Ryan was travelling at a speed of 112 km/h.

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

(a)
From 9.00 a.m. to 1.00 p.m. = 4 h

Distance that Ivan covered when Ryan caught up with Ivan
= 4 x 77
= 308 km

Time that Ryan took to travel 308 km
= 308 ÷ 112
= 2³₄ h

1 h = 60 min
³₄ h = ³₄ x 60 = 45 min
2³₄ h = 2 h 45 min

Time that Ryan left City S:
2 h 45 min before 1.00 p.m. = 10.15 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 539 - 308
= 231 km

Duration that Ryan had to travel before reaching City T
= 231 ÷ 112
= 2¹₁₆ h

Duration that Ivan had to travel before reaching City T
= 231 ÷ 77
= 3 h

Duration that Ivan took to arrive later than Ryan in City T
= 3 - 2¹₁₆
= ¹⁵₁₆ h

Answer(s): (a) 10.15 a.m.; (b) ¹⁵₁₆ h