• Jun 15th 2009, 08:08 AM
foreverbrokenpromises
I buy a number of cheap calculators for a total of $240, mark each of them up by$2 and sell all but four of them for a total of $240. How many calculators did I buy? Okay, so this is what I have done so far: Let y represent the cost per calculator, and let x represent the number of calculator. Therefore, xy = 240 And with the addition of the$2.00, minus the four pencils:
(x-4)(y+2) = 240
xy+2x-4y-8 = 240

and now im stumped. what do i do?

• Jun 15th 2009, 08:22 AM
great_math
Let $x$ be the no. of calculators. Cost per calculator=$$\frac {240}{x}$ $(x-4)\left(\frac {240}{x}+2\right)=240$ $\Rightarrow (x-4)\left(\frac{240+2x}{x}\right)=240$ $\Rightarrow (x-4)(240+2x)=240x$ $\Rightarrow 240x+2x^2-960-8x=240x$ $\Rightarrow 2x^2-8x-960=0$ $\Rightarrow x^2-4x-480=0$ $\Rightarrow (x+20)(x-24)=0$ $\Rightarrow x=-20$(Impossible) or $x=24$ So $24$ is the required answer • Jun 15th 2009, 08:29 AM great_math Quote: Originally Posted by foreverbrokenpromises I buy a number of cheap calculators for a total of$240, mark each of them up by $2 and sell all but four of them for a total of$240. How many calculators did I buy?

Okay, so this is what I have done so far:
Let y represent the cost per calculator, and let x represent the number of calculator.
Therefore, xy = 240
And with the addition of the \$2.00, minus the four pencils:
(x-4)(y+2) = 240
xy+2x-4y-8 = 240

and now im stumped. what do i do?

Now you have to substitute $xy=240$

$240+2x-4y-8=240$

$\Rightarrow 2x-4y=8$

$\Rightarrow x-2y=4$

$\Rightarrow x=4+2y$

Now $xy=240$

$\Rightarrow (4+2y)y=240$

$\Rightarrow 2y^2+4y-240=0$

$\Rightarrow y^2+2y-120=0$

$\Rightarrow (y+12)(y-10)=0$

$\Rightarrow y=-12$(not possible ) or $y=10$

$\Rightarrow x=4+2\times10=24$