Question

A bookstore can purchase several calculators for a total cost of $720. If each calculator cost $1 less, the bookstore could purchase 10 additional calculators at the same total cost. How many calculators can be purchased at the regular price?

Answer 1

Given:

A bookstore can purchase several calculators for a total cost of $720

Let the number of calculators = x

And the cost of one calculator = y

So,

$xy=720\rightarrow(1)$

If each calculator costs $1 less, the bookstore could purchase 10 additional calculators at the same total cost.

so,

$(y-1)(x+10)=720\rightarrow(2)$

divide the equation (2) by equation (1)

$\begin{gathered} \frac{(y-1)(x+10)}{xy}=1 \\ (y-1)(x+10)=xy \\ xy+10y-x-10=xy \\ 10y-x-10=0 \\ x=10y-10\rightarrow(3) \end{gathered}$

Substitute with x into equation (1) then solve for y:

$\begin{gathered} y(10y-10)=720 \\ 10y^2-10y=720 \\ 10y^2-10y-720=0\rightarrow(\div10) \\ y^2-y-72=0 \\ (y+8)(y-9)=0 \\ y+8=0\rightarrow y=-8 \\ y-9=0\rightarrow y=9 \end{gathered}$

Substitute into equation (1) to find x:

$x=\frac{720}{9}=80$

So, the answer will be:

The number of calculators can be purchased at the regular price = 80