Escalators visible steps question. A person always walk down on a moving escalator to save his time. He takes 50 steps while he goes down. One day due to power failure of 10 sec, he took 9 seconds more to get down on than his usual time. What is the number of visible steps of the escalator?
Options are also given...A) 500  B) 450  C) 550  D) None of these.
I am unable to understand the concept of counting visible steps. Is it somewhat related to boats and stream question?
 A: We get an equation system like this from the information provided: 
1) $N = t(V + E)$
2) $N = (t + 9 - 10)E + (t + 9)V = (t - 1)E + (t + 9)V$
Where N is number of steps, E is velocity of escalator (steps/second), V is velocity of person (steps/second) and t is time.
Then we get that $E = 9V$ from these equations. So if person takes $50$ steps normally to complete the distance, escalator moves $9 * 50 = 450$ steps by itself, so the answer is $50 + 450 = 500$ steps. :)
A: One needs one small additional hypothesis for the problem to be well defined: the power failure starts and ends while the person is on the stairs (if not, its effective duration could be less than $10~$seconds, although it logically cannot be less than the $9~$seconds that were lost due to the power failure).
Letting $v_1$ be the number of steps the person takes per unit of time and $v_2$ the number of steps he advances in the same unit of time due to the combined movement of the escalator and her walking. Then while making $50$ steps she advances $\frac{v_2}{v_1}\times50$ steps, which is the number of visible steps on the escalator.
So it suffices to know $\frac{v_2}{v_1}$, and the other information given permits computing it. During the $10~$seconds she advances at speed $v_1$ rather than $v_2$, and apparently this allows the time she passes in combined movement to be reduced by (just) $10-9=1$ second. So $\frac{v_2}{v_1}=10$, and there are $10\times50=500$ (visible) steps in the escalator.
A: Bob enters the staircase at step $-50$ and walks to step $0$ in front of him. Usually he arrives at step $0$ when this step is exactly at the end of the staircase. On the special day he arrives at this step after the usual time, but this step is still $10$ seconds before the end of the staircase. Bob then walks for nine seconds  further ahead to the end of the staircase, doing $r>0$ steps. At this moment step $0$  is $r$ units before the end and reaches the end in one more second. It follows that the escalator is nine times faster in "doing" steps than Bob, and does $450$ steps in the time Bob does $50$ steps.  The length of the staircase therefore is $500$ steps. 
