AAPL wrote:Veritas Prep
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?
$$A.\ \frac{2\left(2c-r\right)}{\left(x-r\right)}$$
$$B.\ \frac{2x\left(c-r\right)}{\left(b-r\right)}$$
$$C.\ \frac{x\left(2c-r\right)}{\left(b-r\right)}$$
$$D.\ \frac{2b\left(c-r\right)}{\left(x-r\right)}$$
$$E.\ \frac{2\left(xc-r\right)}{\left(x-r\right)}$$
OA C
How many umbrellas can it afford to sell at the below-cost rate of b dollars per umbrella and still make a 100% profit?
If c=1 and r=2 -- implying that the cost per umbrella = $1 and that the retail price per umbrella = $2 -- then 100% profit will be yielded only if ALL of the umbrellas are sold at the retail price.
Since this case requires that NONE of the umbrellas be sold at a discount, the correct answer must yield a value of 0 when c=1 and r=2.
Only A and C are guaranteed to yield a value of 0 when c=1 and r=2.
Given c=1 and r=2, it is possible that x=2 umbrellas are sold at the retail price.
In A, x=2 and r=2 will yield a denominator of 0, rendering A invalid.
The correct answer is
C.
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