City Y and City Z is 525 km apart. Tommy left City Y for City Z at 11.00 a.m. travelling at an average speed of 75 km/h. Dylan left City Y later than Tommy and caught up with him at 2.00 p.m. Dylan was travelling at a speed of 100 km/h.

1. At what time did Dylan leave City Y?
2. How much later did Tommy arrive in City Z than Dylan? Express your answer in h as a fraction or mixed number.

(a)
From 11.00 a.m. to 2.00 p.m. = 3 h

Distance that Tommy covered when Dylan caught up with Tommy
= 3 x 75
= 225 km

Time that Dylan took to travel 225 km
= 225 ÷ 100
= 2¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
2¹₄ h = 2 h 15 min

Time that Dylan left City Y:
2 h 15 min before 2.00 p.m. = 11.45 a.m.

(b)
Remaining distance that they had to cover to reach City Z
= 525 - 225
= 300 km

Duration that Dylan had to travel before reaching City Z
= 300 ÷ 100
= 3 h

Duration that Tommy had to travel before reaching City Z
= 300 ÷ 75
= 4 h

Duration that Tommy took to arrive later than Dylan in City Z
= 4 - 3
= 1 h

Answer(s): (a) 11.45 a.m.; (b) 1 h