City Q and City R is 558 km apart. Seth left City Q for City R at 8.00 a.m. travelling at an average speed of 62 km/h. Pierre left City Q later than Seth and caught up with him at 12.00 p.m. Pierre was travelling at a speed of 96 km/h.
- At what time did Pierre leave City Q?
- How much later did Seth arrive in City R than Pierre? Express your answer in h as a fraction or mixed number.
(a)
From 8.00 a.m. to 12.00 p.m. = 4 h
Distance that Seth covered when Pierre caught up with Seth
= 4 x 62
= 248 km
Time that Pierre took to travel 248 km
= 248 ÷ 96
= 2
712 h
1 h = 60 min
712 h =
712 x 60 = 35 min
2
712 h = 2 h 35 min
Time that Pierre left City Q:
2 h 35 min before 12.00 p.m. = 9.25 a.m.
(b)
Remaining distance that they had to cover to reach City R
= 558 - 248
= 310 km
Duration that Pierre had to travel before reaching City R
= 310 ÷ 96
= 3
1148 h
Duration that Seth had to travel before reaching City R
= 310 ÷ 62
= 5 h
Duration that Seth took to arrive later than Pierre in City R
= 5 - 3
1148= 1
3748 h
Answer(s): (a) 9.25 a.m.; (b) 1
3748 h