Let L be the distance between the cities (in some units, miles or kilometers - IT DOES NOT MATTER, as you will see further).

Let u be the speed of the boat in still water (in the CORRESPONDING units), and

let v be the rate of the current (in the same units).


You are given

 = 3  hours    (1)   (rowing downstream)

 = 4  hours    (2)   (rowing   upstream)


From these equations

3*(u+v) = L                 (1')
4*(u-v) = L                 (2')


or, equivalently


3u + 3v = L                 {1'')
4u - 4v = L                 (2'')


What I need is to find the ratio  .  It is EXACTLY the time a piece of driftwood to float downstream from Bowraville to Macksville,
which is under the question.


For it, I am going to eliminate the variable "u" from the equations (1'') and (2'') and to express "v" via L (or, conversely, express L via "v").


So I multiply eq(1'') by 4 (both sides) and multiply eq(2'') by 3 (both sides). I will get

12u - 12v = 3L     (3)
12u + 12v = 4L     (4)


Now subtract eq(3) from eq(4). You will get

24v = L,   which implies   = 24.


It gives the ANSWER:  it will take 24 hours for a piece of driftwood to float downstream from Bowraville to Macksville

(since this piece moves with the rate of the current).