City Q and City R are 460 km apart. At 4.50 p.m., Seth is travelling at a uniform speed left City Q for City R while Xavier set off from City R to City Q along the same road at a uniform speed, which was 11 km/h slower than that of Seth. The two met at 8.50 p.m.
- Find the speed of Seth.
- If Xavier 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 Seth and Xavier to travel from 4.50 p.m. to 8.50 p.m. = 4 h
Average speed of Seth and Xavier
= 460 ÷ 4
= 115 km/h
Seth's speed
= (115 + 11) ÷ 2
= 126 ÷ 2
= 63 km/h
(b)
Xavier's speed
= 63 - 11
= 52 km/h
Distance that Xavier travelled in 4 h
= 4 x 52
= 208 km
Remaining distance that Xavier needed to travel
= 460 - 208
= 252 km
Time that Xavier needed to reach his destination after the two met
= 252 ÷ 52
= 4
4452 = 4
1113 h
Answer(s): (a) 63 km/h; (b) 4
1113 h