SimaQ wrote:

There are 450-5+1=446 floors between them

The travel time of this distance assuming the common rate:

3/4*t=446

t=592 seconds (approximately)

I think the part highlighted is wrong because if A was on the 10th floor and B was on the 5th floor then the amount of distance between them is 5*40

If you draw it:

Even if you're talking about floors then there are

10th---

40ft----9th---

40ft---8th---

40ft---7th---

40ft---6th---

40ft---5th

Now you can see there are five lots of 40ft inbetween the 10th and 5th floor so if we want to calculate the distance in feet it should be (10-5)*40.

Furthermore, if you used the original rationale then the no. of floors worth of distance between the 10th and 9th floor would be (10-9+1)*40 = 80 ---- when the actual distance is 40

I converted everything into floors in order not to mess with feet.... That way it is easier to calculate....

You would have been right if i used the rate of A and B in feet rather than floors.....