SOLUTION: A department store sells 20 portable stereos per week at $80 each. The manager believes that for each decrease of $5 in the price, six more stereos will be sold. What price should

Algebra ->  Equations -> SOLUTION: A department store sells 20 portable stereos per week at $80 each. The manager believes that for each decrease of $5 in the price, six more stereos will be sold. What price should       Log On


   

Question 1121171: A department store sells 20 portable stereos per week at $80 each. The manager believes that for each decrease of $5 in the price, six more stereos will be sold. What price should be charged if the revenue needs to be $2240?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
If n counts each $5 increment
PRICE    QUANTITY SOLD
 80         20
 80-5       20+6
 80-2*5     20+2*6
 80-3*5     20+3*6
.
.
80-5n      20+6n


Revenue to be $2240,
%2880-5n%29%2820%2B6n%29=2240
Simplify and solve.
n is how many $5 increments to remove from $80.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
The condition says that at the price  (80 - 5x) dollars for one stereo the number of stereos sold will be  (20 + 6x),

where x is an integer number.


Then the problem ask to  find "x" in such a way that


    (80-5x)*(20+6x) = 2240.


    1600 - 100x + 480x - 30x^2 = 2240

    - 30x^2 + 380x - 640 = 0

    3x^2 - 38x + 64 = 0


    x%5B1%2C2%5D = %2838+-+sqrt%2838%5E2+-+4%2A3%2A64%29%29%2F%282%2A3%29 = %2838+%2B-+26%29%2F6.


Only positive integer root makes sense, which is  x= 2.


Answer.  The price under the question is  (80-5x) = 80 - 5*2 = 70 dollars.