SOLUTION: I am having a problem fitting my factors into the quadratic equation. The question is as follows:
A retailer spent $48 to purchase a number of special mugs. Two of them were b
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-> SOLUTION: I am having a problem fitting my factors into the quadratic equation. The question is as follows:
A retailer spent $48 to purchase a number of special mugs. Two of them were b
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Question 5014: I am having a problem fitting my factors into the quadratic equation. The question is as follows:
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 remaning mugs for $3 above original cost per mug, she made a total profit of $22.00.
How many mugs were originally purchased?
**If anyone could just give my brain a push, I would be more than grateful!! It's been 15 years since I have used the quadratic equation and I am just stuck!!
Thank you in advance!! Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website! Define: Let x = number of mugs bought.
She broke 2, so she sold (x-2) mugs.
Each mug cost to buy 48/x, but she sold them at 3+(48/x) dollars for a total of 48+22 --> $70.
So, (saleprice per mug)x(number of mugs) = income
(3+(48/x))(x-2) = 70
Multiply both sides by x -->
--> . Collect all terms on left hand side:
This actually factorises to (3x+8)(x-12) = 0 which means that either 3x+8=0 Or x-12=0.
Now, x cannot be -8/3, since x is the number of mugs...-8/3 mugs???, so the answer is she originally bought 12 mugs.
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If you do want to use the quadratic formula:
x =
where
a = 3
b = -28
c = -96
--> x =
--> x =
--> x =
--> x =
so either x = or x =
