emmak wrote:

A manufacturer makes umbrellas at the cost of c dollars per umbrella, and sells them at the retail price of r dollars each. How many umbrellas can it afford to sell at the below-cost rate of b dollars per umbrella and still make a 100% profit on its lot of x total umbrellas?

A. \(\frac{b(2c-r)}{(x-r)}\)

B. \(\frac{2x(c-r)}{(b-r)}\)

C. \(\frac{x(2c-r)}{(b-r)}\)

D. \(\frac{2b(c-r)}{(x-r)}\)

E. \(\frac{2(xc-r)}{(x-r)}\)

Official solution from

Veritas Prep.

Correct Answer: C

For this problem, it is important to note that \(Profit = Revenue - Cost\). The profit needs to be equal the cost in order to make a 100% profit, so essentially the equation should then be \(Cost = Revenue - Cost\). The cost to the manufacturer is the number of units it makes, x, multiplied by the cost per unit, c, for a product of \(xc\). The revenue that the manufacturer will bring in has two components: the number it sells at the sale price of b multiplied by that price, b, and the rest of the umbrellas, which it sells at r, times r. If we call the number it is allowed to sell at a discount y, then we can solve for y to get our answer. That would make the revenue \(yb + (x-y)r\).

Therefore, the equation Cost = Revenue - Cost will be:

\(xc = yb + (x-y)r - xc\).

Algebraically, we need to manipulate this equation to solve for y:

\(xc = yb + (x-y)r - xc\)

\(xc = yb + xr - yr - xc\) distribute the multiplication to get y out of the parentheses

\(2xc = yb + xr - yr\) add xc to both sides

\(2xc - xr = yb - yr\) subtract xr from both sides to get the y terms on their own

\(2xc - xr = y(b - r)\) factor the y to get it on its own

\(\frac{(2xc−xr)}{(b−r)} = y\) divide both sides by \((b - r)\) to get y on its own

\(x \frac{(2c−r)}{(b−r)} = y\) factor the x to make this equation look like the answer choices.

Because the solution for y matches answer choice C, the answer is C.

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