Comments on: How many comic books were in the box? http://www.thecliffguy.com/comic-books/how-many-comic-books-were-in-the-box Comic Book Information Wed, 08 Sep 2010 22:11:11 +0000 http://wordpress.org/?v=2.5.1 By: Ron W http://www.thecliffguy.com/comic-books/how-many-comic-books-were-in-the-box#comment-1992 Ron W Mon, 02 Nov 2009 23:59:59 +0000 http://www.thecliffguy.com/comic-books/how-many-comic-books-were-in-the-box#comment-1992 Let x denote the number of comic books in the box. Her cost per magazine was 300/x. If she gave 15 to her brother, then she sold (x-15) magazines on the internet site. The price she got per magazine was 330/(x-15). This was $1.50 more than her cost per magazine. So 330/(x-15) - 300/x = 1.5 Multiply both sides by x(x-15): 330x - 300(x-15)) = 1.5x(x-15) 330x - 300x + 4500 = 1.5x² - 22.5x 30x + 4500 = 1.5x² - 22.5x Divide both sides by 1.5 (I like integers) 20x + 3000 = x² - 15x 0 = x² - 35x - 3000 Using the quadratic formula, x = [35 ± √(1225 + 12000)]/2 x = 75 or x = -40 We must reject the negative solution of course. The original price of the comic books, per book, was $300/75 = $4<br><b>References : </b><br> Let x denote the number of comic books in the box. Her cost per magazine was 300/x. If she gave 15 to her brother, then she sold (x-15) magazines on the internet site. The price she got per magazine was 330/(x-15). This was $1.50 more than her cost per magazine. So

330/(x-15) - 300/x = 1.5

Multiply both sides by x(x-15):

330x - 300(x-15)) = 1.5x(x-15)
330x - 300x + 4500 = 1.5x² - 22.5x
30x + 4500 = 1.5x² - 22.5x

Divide both sides by 1.5 (I like integers)

20x + 3000 = x² - 15x
0 = x² - 35x - 3000

Using the quadratic formula,

x = [35 ± √(1225 + 12000)]/2
x = 75 or x = -40

We must reject the negative solution of course.

The original price of the comic books, per book, was $300/75 = $4
References :

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