Interesting physics problem about escalator

Question

A physics student decided to investigate the escalator. By the time he went down the escalator for the first time, he counted n₁ = 50 steps. The second time, moving down at three times the speed, he counted n₂ = 75 steps.

In which direction does the escalator move? How many steps would he count if the escalator did not move?

Answer 1

Let u steps/s is speed of escalator, v steps/s is speed of student. Let escalator moves up.

N steps is length of escalator.

We have: N·v/(u + v) = 50; N·3v/(u + 3v) = 75.

By division we get (u + 3v)/(u + v) = 2; u + 3v = 2u + 2v; v = u

Then Nu/(u + u) = 50; N/2 = 50; N = 100 steps.

If we suppose that escalator moves down: vn/(v-u) = 50 and 3Nv/(3v - u) = 75; (3v - u)/(v - u)= 2; 3v - u = 2v - 2u; v = - u. This is contradiction.