A computer store manager buys several computers of the same model for $7,600. The store can regain this investment by selling all but 9 of the computers at a profit of $360 per computer. To do this, how many computers must be sold, and at what price?

Question

Mathematics

Spencer Carr

11 March, 04:22

A computer store manager buys several computers of the same model for $7,600. The store can regain this investment by selling all but 9 of the computers at a profit of $360 per computer. To do this, how many computers must be sold, and at what price?

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Answers (1)

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Amaris Mullen · Answer 1

Let,

he bought x computers for $7600

so he bought one for 7600/x$

as he sold the computers for a $360 profit each,

he sold a computer for (7600/x + 360) $

= (7600+360x) / x $

as all but nine computers together covered up $7600,

we can say that,

(x-9) (7600+360x/x) = 7600

x (7600+360x/x) - 9 (7600+360x/x) = 7600

7600+360x - 9 (7600+360x/x) = 7600

360x-9 (7600+360x/x) = 0

360x=9 (7600+360x/x)

40x=7600+360x/x

x=190+9x/x

x (squared) = 190+9x

x (s ...) - 9x-190=0

x (s ...) - 19x+10x - 190=0

x (x-19) + 10 (x-19) = 0

(x-19) (x+10) = 0

x=19 (x must be a positive quantity so x+10 cancels out)

so he sold 19-9=10 computers