# Precalculus question word problem? [closed]

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.

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## closed as off-topic by Thursday, Fly by Night, Babak S., rogerl, Sami Ben RomdhaneAug 3 at 22:52

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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

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 $S=70$.