Maths help...

[cshelmo]

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 for $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.

a. If the price for n mugs is $48, how can we express the cost per mug?

b. We are told that the retailer, in order to make a specific profit, sells the remiaining mugs for $3 more than the original cost per mug. Using this information, and the answer to step a, how can we express the selling price per mug?

c. Since two of the mugs were broken in the store, how can we represent the number of mugs that are available for sale?

d. Use the answers from step b and c to construct an expression for the amount of money made on the sale of the remaining mugs.

e. The profit on the mugs can be found by subtracting the cost of the mugs from the amount of money made on the slae of the mugs. Write an equation that states this relationship between money made, the cost, and the profit.

[OBonio]

a) 48/n
b) (48/n) + 3
c) n - 2
d) (n-2)((48/n) + 3)
e) (n-2)((48/n) + 3) - 48 = 22

(n-2)(48/n + 3) = 70
48 + 3n -96/n +3n = 70
48n + 3n^2 -96 + 3n^2 = 70n

6n^2 - 22n = 96

Err.. then use that thing to solve quadratic equations :o)

[OBonio]

Hmm... didn't expand my parentheses too well there..

[OBonio]

48/n(n-2) + 3(n-2) = 70
48 - 96/n + 3n - 6 = 70
-96/n +3n = 70 + 6 - 48
-96/n + 3n = 28
-96 + 3n^2 = 28n
3n^2 - 28n = 96

Then use that quadratic thingymabob.

[gen2]

You don't need the thingymabob, just rearrange the terms and factorise.

3n^2 - 28n = 96
3n^2 - 28n - 96 = 0
(3n + 8) ( n - 12) = 0

so either (3n + 8) = 0 ( ==> n = -8/3)
or ( n - 12) = 0 ( ==> n = 12)

[OBonio]

To corroborate gen2's answer :

n = (-b +/- (b^2 - 4ac)^(1/2)) / 2a
n = (28 +/- (784 + 1152)^(1/2)) / 6
n= (28 +/- 44) / 6
Therefore :
n = -16/6 = -8/3 = -2 and 2 thirds
or
n = 72/6 = 12

I think the 2nd one to be more likely.

[Fink-Knottle]

Que?