City Y and City Z is 408 km apart. Tom left City Y for City Z at 11.00 a.m. travelling at an average speed of 68 km/h. Gabriel left City Y later than Tom and caught up with him at 1.00 p.m. Gabriel was travelling at a speed of 96 km/h.

1. At what time did Gabriel leave City Y?
2. How much later did Tom arrive in City Z than Gabriel? 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 Tom covered when Gabriel caught up with Tom
= 2 x 68
= 136 km

Time that Gabriel took to travel 136 km
= 136 ÷ 96
= 1⁵₁₂ h

1 h = 60 min
⁵₁₂ h = ⁵₁₂ x 60 = 25 min
1⁵₁₂ h = 1 h 25 min

Time that Gabriel left City Y:
1 h 25 min before 1.00 p.m. = 11.35 a.m.

(b)
Remaining distance that they had to cover to reach City Z
= 408 - 136
= 272 km

Duration that Gabriel had to travel before reaching City Z
= 272 ÷ 96
= 2⁵₆ h

Duration that Tom had to travel before reaching City Z
= 272 ÷ 68
= 4 h

Duration that Tom took to arrive later than Gabriel in City Z
= 4 - 2⁵₆
= 1¹₆ h

Answer(s): (a) 11.35 a.m.; (b) 1¹₆ h