Question

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?

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