SOLUTION: The word problem is....A retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store but by sellimg each of the remaining mugs at $3 above the ori
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Question 458230: The word problem is....A retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store but by sellimg each of the remaining mugs at $3 above the original cost per mug she made a total profit of $22. Construct an equation that will allow us to solve for the number of mugs, denoted by n, that were originally purchased.
If the price for n mugs is $48, how can we express the cost per mug?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A retailer spent $48 to purchase a number of special mugs.
Two of them were broken in the store but by selling each of the remaining mugs
at $3 above the original cost per mug she made a total profit of $22.
Construct an equation that will allow us to solve for the number of mugs, denoted by n, that were originally purchased.
If the price for n mugs is $48, how can we express the cost per mug?
:
Price paid for each mug =
:
No. of mugs sold = (n-2)
:
Selling price = + 3
;
Revenue should = cost + profit
:
Total revenue = 48 + 22 = $70:
:
#sold * Price sold = total revenue
(n-2) * [ + 3] = 70
FOIL
48 + 3n - 96/n - 6 = 70
combine like terms
3n - 96/n + 48 - 6 - 70 = 0
3n - 96/n - 28 = 0
Multiply equation by n, forming a quadratic equation:
3n^2 - 28n - 96 = 0
Factor
(3n +8)(n-12) = 0
Positive solution
n = 12 mugs originally bought
:
Check
Cost: 48/12 = $4
Sold: 10 * 7 = $70 (cost + profit)
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