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)