SOLUTION: A retailer bought a number of special mugs for $48. She decided to keep two of the mugs for herself but then had to change the price to $ 3 a mug above the original cost per mug.

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Question 188523This question is from textbook
: A retailer bought a number of special mugs for $48. She decided to keep two of the mugs for herself but then had to change the price to $ 3 a mug above the original cost per mug. If she sells the remaining mugs for $70, how many mugs didi she buy and at what price per mug did she sell them?
Thanks This question is from textbook

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A retailer bought a number of special mugs for $48. She decided to keep two
of the mugs for herself, but then had to change the price to $3 a mug above
the original cost per mug. If she sells the remaining mugs for $70,
how many mugs did she buy and at what price per mug did she sell them?
:
Let n = number of mugs she bought
then
(n-2) = number of mugs sold (she kept two)
:
48%2Fn = original cost per mug
:
70%2F%28%28n-2%29%29 = price per mug she sold
:
Original cost + $3 = price per mug she sold
48%2Fn + 3 = 70%2F%28%28n-2%29%29
Multiply n(n-2) to get rid of the denominators, results:
48(n-2) + 3n(n-2) = 70n
:
48n - 96 + 3n^2 - 6n = 70n
Arrange as a quadratic equation
3n^2 + 48n - 6n - 70n - 96 = 0
:
3n^2 - 28n - 96 = 9
Factor
(3n+8)(n-12) = 0
The positive solution is what we want here
n = 12 mugs bought
then
10 mugs sold
:
70%2F10 = $7 price of each mug sold
;
:
Check solution
70/10 = $7 price
48/12 = $4 cost
-----------------
Mark-up=$3
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