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A boy is climbing up a flight of $35$ steps from Level $1$ to Level $2$ of a building. If he climbs at his fastest speed for $15$ steps and slowest speed for $15$ steps, he will take $40$ seconds to climb $30$ steps. If he climbs at his fastest speed for $26$ steps and slowest speed for $9$ steps, he will reach Level $2$ from Level $1$ in $41$ seconds. Find the shortest time it will take him to reach Level $2$ from Level $1$.

The conditions for the first given hint and the second given hint are different - one is for $30$ steps total and the other one is for $35$ steps. How would I solve this?

Darío A. Gutiérrez
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2 Answers2

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Set variables. Let his slowest speed be $x$ seconds per step and let his fastest speed be $y$ seconds per step. Then try to turn the pieces of data you have into equations.

Carl Schildkraut
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Think of it this way: set his slowest speed as $x$ seconds per step, and his fastest speed as $y$ seconds per step. Then you can get two equations:

$(1)\,15\,\mathrm{steps} \cdot x\,\mathrm{\frac{secs}{step}}+15\,\mathrm{steps} \cdot y\,\mathrm{\frac{secs}{step}}=40\,\mathrm{secs}$
$(2)\,26\,\mathrm{steps} \cdot x\,\mathrm{\frac{secs}{step}}+9\,\mathrm{steps} \cdot y\,\mathrm{\frac{secs}{step}}=41\,\mathrm{secs}$
which cancels to
$(1)\,15x\,\mathrm{secs}+15y\,\mathrm{secs}=40\,\mathrm{secs}$
$(2)\,26x\,\mathrm{secs}+9y\,\mathrm{secs}=40\,\mathrm{secs}$
then, dividing the $\mathrm{secs}$ unit from each side
$(1)\,15x+15y=40$
$(2)\,26x+9y=41$

Can you solve from here? As you can see, the "different" amount of steps was merely an illusion. Generally you can break down system of equations problems questions like this using unit cancellation.

ericw31415
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