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A manufacturer makes umbrellas at the cost of c dollars per

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by AAPL » Tue Feb 26, 2019 4:35 am

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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
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by Brent@GMATPrepNow » Tue Feb 26, 2019 5:59 am
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

This is a tough one to use the INPUT-OUTPUT approach, but here is goes:

Let c = $2 (it cost $2 to make each umbrella)
Let x = 10 (we make 10 umbrellas)
Let r = $5 (the retail price is $5 per umbrella)
Let b = $0 (the below-cost sale price is $0 per umbrella)

So, the manufacturer made 10 umbrellas at the cost of $2 per umbrella. So the total cost = $20
We need a 100% profit. So, we must earn $40 in revenue. In other words, we must sell 8 umbrellas at $5 each.
This means we can "sell" 2 umbrellas at the below-cost sale price of $0 each.

At this point, we must plug c = 2, x = 10, r = 5 and b = 0 into each expression and see which one yields an OUTPUT of 2

A. 2(2c-r)/(x-r) = -2/5 ELIMINATE A

B. 2x(c-r)/(b-r) = 12 ELIMINATE B

C. x(2c-r)/(b-r) = 2 KEEP C

D. 2b(c-r)/(x-r) = 0 ELIMINATE D

E. 2(xc-r)/(x-r) = 6 ELIMINATE E

Answer : C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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umbrellas

by GMATGuruNY » Tue Feb 26, 2019 7:19 am
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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by Scott@TargetTestPrep » Fri Mar 01, 2019 9:35 am
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
Let y = the number of umbrellas the manufacturer sold below cost for b dollars each. Thus, by is the revenue from the below-cost part of the sale.

Since x is the total number of umbrellas sold, then he sold x - y umbrellas for r dollars each. Thus, r(x - y) is the revenue from the above-cost part of the sale. Since we want to have 100% profit, the revenue R has to be twice the cost:

R = 2c

by + r(x - y) = 2cx

by + rx - ry = 2cx

by - ry = 2cx - rx

y(b - r) = 2cx - rx

y = (2cx - rx)/(b - r)

y = x(2c - r)/(b - r)

Answer: C

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