URGENT MATHS HELP NEEDED! Please help?

Question

A group of hikers is to travel x km by bus at an average speed of 48 km/h to an unknown destination. They then plan to walk back along the same route at an average speed of 4.8 km/h and to arrive back 24 hours after setting out in the bus. If they allow 2 h for lunch and rest, how far must the bus take them?

Help would be appreciated.. I really dont know where to begin. Any help would be fab!

Answer 1

d = r*t
48t = (22-t)4.8
48t = 105.6 -4.8t
43.2t = 105.6
t = 2 4/9 hours
d = 2 4/9 * 48 = 117 1/3 km

Answer 2

they are on the bus for 2 hours, and walking for 20 hours

let's say that they are on the bus for t hours, then they are walking for 22-t hours (of the 24 hours of the entire trip, 2 are spent not traveling)

recall that distance = speedxtime, so we can write:

the distance traveled by bus is then 48t, and the distance traveled on foot is 4.8(22-t)

equate 48t=4.8(22-t)
solving this gives t=2 hours; so they travel 96 km on the bus, and also 96 km (20 hoursx4.8km/hr) on foot

Another way to see this is that since they are traveling 10 times faster in the bus than on foot, they must spend ten times more time on foot. If the total travel time is 22 hours, then they must spend 20 hours on foot, and 1/10 of that, 2 hours on the bus

Answer 3

Well distance walk = distance bus:

Dw = Db

Distance bus = time bus * 48
also
Distance walk = time walk * 4.8
Which means
4.8 * time walk = 48 * time bus
time walk / time bus = 10
But they allow 2 hours break so

Time walk + Time bus + 2 = 24.

So

time walk + time bus = 22

but also
time walk / time bus = 10

2 hours on the bus
20 hours walking
2 * 48 or 20 * 4.8

:3