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Answer 1 · 2017-06-16T12:47:19

I got $15$ $15$ pens at $$ 6$ $$6$ each.

Call the number of pens $x$ $x$ and the cost of one pen $y$ $y$ we get:

At the original price:
$x y = 90$ $xy=90$

If we subtract $$ 1$ $$1$ from the original price we can buy $3$ $3$ more pens or:
$(x + 3) (y - 1) = 90$ $(x+3)(y-1)=90$

so
$(x + 3) (\frac{90}{x} - 1) = 90$ $(x+3)(90/x-1)=90$
$90 - x + \frac{270}{x} - 3 = 90$ $90-x+270/x-3=90$
$- x^{2} + 270 - 3 x = 0$ $-x^2+270-3x=0$

Using the quadratic Formula we get that:

$x_{1} = 15$ $x_1=15$ (we get a negative solution as well $x_{2} = - 18$ $x_2=-18$ ).

So the total number of pens is $15$ $15$ and the cost of one pen will be:
$y = \frac{90}{x} = \frac{90}{15} = $ 6$ $y=90/x=90/15=$6$