City Y and City Z are 595 km apart. At 1.43 p.m., Paul is travelling at a uniform speed left City Y for City Z while Seth set off from City Z to City Y along the same road at a uniform speed, which was 7 km/h slower than that of Paul. The two met at 6.43 p.m.
- Find the speed of Paul.
- If Seth continued to travel at the same speed, how long would it take for him to reach his destination after the two met? Express your answer in mixed number.
(a)
Time taken for Paul and Seth to travel from 1.43 p.m. to 6.43 p.m. = 5 h
Average speed of Paul and Seth
= 595 ÷ 5
= 119 km/h
Paul's speed
= (119 + 7) ÷ 2
= 126 ÷ 2
= 63 km/h
(b)
Seth's speed
= 63 - 7
= 56 km/h
Distance that Seth travelled in 5 h
= 5 x 56
= 280 km
Remaining distance that Seth needed to travel
= 595 - 280
= 315 km
Time that Seth needed to reach his destination after the two met
= 315 ÷ 56
= 5
3556 = 5
58 h
Answer(s): (a) 63 km/h; (b) 5
58 h