City R and City S is 693 km apart. Xavier left City R for City S at 11.00 a.m. travelling at an average speed of 77 km/h. Jeremy left City R later than Xavier and caught up with him at 3.00 p.m. Jeremy was travelling at a speed of 112 km/h.
- At what time did Jeremy leave City R?
- How much later did Xavier arrive in City S than Jeremy? Express your answer in h as a fraction or mixed number.
(a)
From 11.00 a.m. to 3.00 p.m. = 4 h
Distance that Xavier covered when Jeremy caught up with Xavier
= 4 x 77
= 308 km
Time that Jeremy took to travel 308 km
= 308 ÷ 112
= 2
34 h
1 h = 60 min
34 h =
34 x 60 = 45 min
2
34 h = 2 h 45 min
Time that Jeremy left City R:
2 h 45 min before 3.00 p.m. = 12.15 p.m.
(b)
Remaining distance that they had to cover to reach City S
= 693 - 308
= 385 km
Duration that Jeremy had to travel before reaching City S
= 385 ÷ 112
= 3
716 h
Duration that Xavier had to travel before reaching City S
= 385 ÷ 77
= 5 h
Duration that Xavier took to arrive later than Jeremy in City S
= 5 - 3
716= 1
916 h
Answer(s): (a) 12.15 p.m.; (b) 1
916 h