"A train leaves Newcastle for Sydney at 7:45am while another train leaves..."

sntnl

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"A train leaves Newcastle for Sydney at 7:45am while another train leaves..."

A train leaves Newcastle for Sydney at 7:45am while another train leaves from Sydney for Newcastle at 8:15am. The first train arrives in Sydney 40 minutes after the trains pass while the train to Newcastle arrives 1 hour 40 minutes after they pass. Assuming that the trains travel at constant speeds (not necessarily the same), at what time did they pass?

Not sure what topic to put this under, as I have not solved it. Taken from a year 8 book.
 
A train leaves Newcastle for Sydney at 7:45am while another train leaves from Sydney for Newcastle at 8:15am. The first train arrives in Sydney 40 minutes after the trains pass while the train to Newcastle arrives 1 hour 40 minutes after they pass. Assuming that the trains travel at constant speeds (not necessarily the same), at what time did they pass?

Not sure what topic to put this under, as I have not solved it. Taken from a year 8 book.
What are your thoughts?

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Had a look at the solutions at other sites;
they seem quite difficult to follow.
Here's the way I'd solve:
(X=Newcastle, Y = Sydney, A = train1, B = train2)

Simplify problem:
A leaves X at time 0, travelling to Y, at speed a.
B leaves y 30 minutes later, travelling to A, at speed b. If Y is Sydney, where is 'y' - Tasmania??!! I think LA has infected me .....
A meets B t minutes later.
[X](A){0}@a.......................>t<.........................@b{30}(B)[Y]

A continues and arrives at Y 40 minutes later.
B continues and arrives at X 100 minutes later.
[X](B){40}<..................@b(B),(A)@a..................>{100}(A)[Y]

d = difference in starting time (30)
u = time by A after meeting (40)
v = time by B after meeting (100)
t = time by A to meet B (?)

t = [d + SQRT(d^2 + 4uv)] / 2
That'll give you t = 80 minutes.

Now add that to actual 7.45am to get 9.05am

Notice that the speeds and distances are not required.

A real nice problem!!
.
 
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