City B and City C is 560 km apart. Peter left City B for City C at 7.00 a.m. travelling at an average speed of 70 km/h. Vaidev left City B later than Peter and caught up with him at 11.00 a.m. Vaidev was travelling at a speed of 112 km/h.

1. At what time did Vaidev leave City B?
2. How much later did Peter arrive in City C than Vaidev? Express your answer in h as a fraction or mixed number.

(a)
From 7.00 a.m. to 11.00 a.m. = 4 h

Distance that Peter covered when Vaidev caught up with Peter
= 4 x 70
= 280 km

Time that Vaidev took to travel 280 km
= 280 ÷ 112
= 2¹₂ h

1 h = 60 min
¹₂ h = ¹₂ x 60 = 30 min
2¹₂ h = 2 h 30 min

Time that Vaidev left City B:
2 h 30 min before 11.00 a.m. = 8.30 a.m.

(b)
Remaining distance that they had to cover to reach City C
= 560 - 280
= 280 km

Duration that Vaidev had to travel before reaching City C
= 280 ÷ 112
= 2¹₂ h

Duration that Peter had to travel before reaching City C
= 280 ÷ 70
= 4 h

Duration that Peter took to arrive later than Vaidev in City C
= 4 - 2¹₂
= 1¹₂ h

Answer(s): (a) 8.30 a.m.; (b) 1¹₂ h