City R and City S is 380 km apart. Bryan left City R for City S at 8.00 a.m. travelling at an average speed of 76 km/h. Howard left City R later than Bryan and caught up with him at 10.00 a.m. Howard was travelling at a speed of 96 km/h.

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

(a)
From 8.00 a.m. to 10.00 a.m. = 2 h

Distance that Bryan covered when Howard caught up with Bryan
= 2 x 76
= 152 km

Time that Howard took to travel 152 km
= 152 ÷ 96
= 1⁷₁₂ h

1 h = 60 min
⁷₁₂ h = ⁷₁₂ x 60 = 35 min
1⁷₁₂ h = 1 h 35 min

Time that Howard left City R:
1 h 35 min before 10.00 a.m. = 8.25 a.m.

(b)
Remaining distance that they had to cover to reach City S
= 380 - 152
= 228 km

Duration that Howard had to travel before reaching City S
= 228 ÷ 96
= 2³₈ h

Duration that Bryan had to travel before reaching City S
= 228 ÷ 76
= 3 h

Duration that Bryan took to arrive later than Howard in City S
= 3 - 2³₈
= ⁵₈ h

Answer(s): (a) 8.25 a.m.; (b) ⁵₈ h