Help with a maths question ?

Let 2d km be the distance between the market and Debbie's home.

Time taken = Distance/Speed
Time taken for Debbie to meet her mother = (d - 140)/55 min
Time taken for her mother to meet Debbie = (d + 140)/90 min

(d - 140)/55 = (d + 140)/90
90 × (d - 140) = 55 × (d + 140)
90d - 12600 = 55d + 7700
35d = 20300
d = 580
2d = 1160

Answer: The market was 1160 km from Debbie's home.

distance = rate * time

We can set up two equations, one for Debbie and one for her mother:

d = rt and D = RT

We are told what their rates are:

r = 55 and R = 90

Their times are the same:

t = T

And they met 140 m from the middle.  This is the part that takes a little bit to think about and may need to be sketched to get a good look at how to set this up.

I drew a line from start to finish and called it "x", which is what the question is asking us to find.  We know that this will be the sum of both distances, so:

d + D = x

The middle of the path is then (x/2). D is then 140m more than x/2 if that's where they met:

D = x/2 + 140

So now we have 7 equations and 7 unknowns, so we can now start substituting pieces of information into place and eventually solve for "x".

Starting with these equations:

d = rt and D = RT and d + D = x and D = x/2 + 140

Let's first substitute our constants and other variables r, R, and T:

d = 55t and D = 90t and d + D = x and D = x/2 + 140

Now we have "d" and "D" in terms of "t".  Substitute those into the final two equations, then simplify.  The second equation I'll multiply both sides by 2 in this step to get rid of the fraction:

d + D = x and D = x/2 + 140
55t + 90t = x and 90t = x/2 + 140
145t = x and 180t = x + 280

We have "x" in terms of "t", substitute into the last equation to solve for "t":

180t = x + 280
180t = 145t + 280
35t = 280
t = 8

Now that we have "t", we can solve for "x":

x = 145t
x = 145(8)
x = 1160

There was an 1160 m distance between the house and the market.