An Oxford Entrance Math Exam Question

I think it’s like this (I could be wrong):

Let t = time taken by the “blue” car when the two cars meet (in hours).

And, d = distance travelled by both cars when they meet (in km).

So, the invariant here is “d”.

Thus since

Distance = Velocity x Time,

We have the distance travelled,

(i) For the red car

d = 40 x (t + 3/60)

(ii) For the blue car

d = 60 x t

Then equating the above and solving,

60t = 40 (t + 3/60)
60t = 40t + 2
20t = 2
t = 2/20
Or,
t = 1/10 hours = 6 minutes

Put back to calculate the distance travelled by the red car,

d = 40 x (1/10 + 3/60)
d = 40 x (1/10 + 1/20)
d = 40 x (3/20)
d = 2 x 3
d = 6 kilometres (Answer)