Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Leon who is always in a hurry walked up an escalator while it was moving at the rate of one step per second and reached the top in 30 steps. the next day he climbed two steps per second also while it was moving and reached the top in 42 steps.If the escalator had been stopped how many steps did the escalator have from the bottom to the top.

share|improve this question
    
It seems to me that your first sentence can be construed to mean that the escalator was moving at the rate of one step per second. –  David Mitra Apr 13 '12 at 16:45

1 Answer 1

I will assume that it is Leon who on the first day was walking at one step per second (kind of slow, really), and not the elevator.

Let $S$ be the number of steps on the escalator, and let $v$ be the velocity of the accelerator, in steps per second.

On the first day, our hero took $30$ steps, and in the $30$ seconds it took, the escalator lifted him by a further $30v$ steps. It follows that
$$S=30+30v.$$ On the second day, going up took $21$ seconds, so the same kind of reasoning leads to the equation $$S=42+21v.$$ Solve the system. It is reasonably efficient to multiply each side of the first equation through by $7$, giving $7s=210+210v$. In the second equation, multiply through by $10$, getting $10S=420+210v$. We conclude that $3S=210$, and therefore $S=70$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.