Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q television sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions? (1) The Model P televisions sold for $30 less than the Model Q televisions. (2) Either p = 120 or q = 120.

Let [tex]x[/tex] be the number of Model P sold and [tex]y[/tex] the number of Model Q sold. This problem has 4 unknown variables. With the information on the problem you can write some equations:

The total of television sold was 40:

[tex]x+y=40[/tex]

The average of the selling price was $141:

[tex]\frac{px+qy}{40}=141[/tex]

The Model P sold for $30 less than the other model:

[tex]p+30=q[/tex]

With only three equation, is needed to test them with q=120 and p=120

if [tex]q=120[/tex]

[tex]p+30=q\\p+30=120\\p=120-30=90[/tex]

[tex]\frac{90x+120y}{40}=141\\2.25x+3y=141[/tex]

[tex]x+y=40[/tex]

Solving the system of equation by the method of elimination (Multiply this equation [tex]x+y=40[/tex] by -3):

[tex]-3x-3y=-120\\2.25x+3y=141\\---------\\-0.75x+0=21\\x=-21/0.75\\x=-28[/tex]

Substitute the value of [tex]x[/tex] in one of the equations:

[tex]x+y=40\\-28+y=40\\y=40+28\\y=68\\[/tex]

With a [tex]y[/tex] greater than 40 and a negative value of [tex]x[/tex], this can't be the solution.

if [tex]p=120[/tex]

[tex]p+30=q\\120+30=q\\q=150[/tex]

[tex]\frac{120x+150y}{40}=141\\3x+3.75y=141\\[/tex]

Solving the system of equation by the method of elimination (Multiply this equation [tex]x+y=40[/tex] by -3):

[tex]-3x-3y=-120\\3x+3.75y=141\\---------\\0+0.75y=21\\0.75y=21\\y=21/0.75\\y=28[/tex]

Substitute the value of [tex]y[/tex] in one of the equations:

[tex]x+y=40\\x+28=40\\x=40-28\\x=12[/tex]

They sold 12 Model P televisions and 28 Model Q televisions.