A li'l travelling saga I made up for your pleasure!!

X(A,B,C)..................2952 miles....................Y

XY is a straight road, length 2952 miles.
A, B and C are together at X, and leave at the same time.
A's speed is 1 mph, B's speed is 2 mph, C's speed is 8 mph.

C travels to Y, then returns and meets A.
He picks up A and returns towards Y, at C's speed.
He drops off A at some distance from Y; A continues.

C returns and meets B.
He picks up B and returns towards Y, at C's speed.
C and B arrive at Y at same time as A does.

After how many hours do they arrive at Y?

I know that by the time C picks up A he has travelled for 369 hours getting to Y at 8 mph and another 287 hours going back to meet A at a combined speed of 9 mph, also that this will leave them both 1640 miles from Y.

B will meanwhile have covered 1312 miles at 2 mph.

Using S as the distance from Y where C leaves A and R as the distance from Y where C picks up B (travelling back to meet him at a combined speed of 10 mph), I have an equation along the lines of S = (R-S)/10 [simplified from S + R/8 = (R-S)/10 + R/8] for the unknown part of the journey time, and I think possibly R = 1640 - (R-S)/10 although I am less sure of that and in any event would struggle to make any more progress...

This sort of maths was all too long ago for me!

[X]C@8......................2952.....................>369h[Y]

[X]A@1---656--->656h<...........2296...........C@8[Y]

Good try Lex, but 1640 is incorrect; should be 2296.
When they meet 656 hours later, A has travelled 656 miles.

I think that at one point I had juxtaposed the figures and had 2592.

[X]A@1---656--->656h<...........2296...........C@8[Y]

Following this point, we have 3 "events":
(w, x and y represents hours so far)

1: C picks up A, returns, drops off A after w hours;
A continues and arrives at Y after y hours.
Equation: 656 + 8(w - 656) + 1(y - w) = 2952

2: C returns and picks up B after x hours.
Equation: 2x + 8(x - w) + 1(y - w) = 2952

3: C returns and arrives at Y after y hours.
Equation: 2x + 8(y - x) = 2952

So we have 3 equations and 3 unknowns (w,x,y).

Too many unknowns for me: I don't think my additional maths O Level (in about 1965) took me beyond 2 unknowns...

A bit "longish", but not difficult:

http://www.sosmath.com/soe/SE311105/SE311105.html