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# A manufacturer makes umbrellas at the cost of c dollars pe

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A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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27 Feb 2013, 04:16
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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. $$\frac{2(xc-r)}{(x-r)}$$
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Feb 2013, 05:53, edited 1 time in total.
Edited the question.
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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27 Feb 2013, 04:27
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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?

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

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

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

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

$$\frac{2(xc-r)}{(x-r)}$$

Let "N" be the required answer.,

(x - N)r + Nb = 2xc

xr - Nr + Nb = 2xc

N(b - r) = x(2c - r)

N = $$\frac{x(2c-r)}{(b-r)}$$

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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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28 Feb 2013, 03:33
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process of elimination + intuition
x and r are not of the same nature (cost in dollar and total umbrellas) => eliminate A,D,E
now, with some intuition, if retail price is less than cost, B would be negative => eliminate B
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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11 Jan 2014, 08:36
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. $$\frac{2(xc-r)}{(x-r)}$$

Could anybody try smart numbers on this one? I'm having a tough time coming up with the solution

Thanks!
Cheers!
J
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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13 Sep 2014, 16:44
leekay wrote:
process of elimination + intuition
x and r are not of the same nature (cost in dollar and total umbrellas) => eliminate A,D,E
now, with some intuition, if retail price is less than cost, B would be negative => eliminate B

Good approach, but still I don´t understand how to choose between B and C.
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22 Oct 2015, 08:59
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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22 Oct 2015, 11:39
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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. $$\frac{2(xc-r)}{(x-r)}$$

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 x = 2, x = 10, r = 5 and b = 10 into each expression and see which one yields an OUTPUT of 2

A. $$\frac{b(2c-r)}{(x-r)}$$ = 0 ELIMINATE A

B. $$\frac{2x(c-r)}{(b-r)}$$ = 12 ELIMINATE B

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

D. $$\frac{2b(c-r)}{(x-r)}$$ = 0 ELIMINATE D

E. $$\frac{2(xc-r)}{(x-r)}$$ = 6 ELIMINATE E

Cheers,
Brent
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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22 Oct 2015, 11:40
MacFauz and I have demonstrated the two methods (Algebraic and Input-Output) for solving a question type I call Variables in the Answer Choices.

If you'd like more information on these approaches, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935

Cheers,
Brent
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Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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16 Jan 2017, 03:16
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A manufacturer makes umbrellas at the cost of c dollars pe [#permalink]

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28 Feb 2017, 22:12
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.

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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A manufacturer makes umbrellas at the cost of c dollars pe   [#permalink] 28 Feb 2017, 22:12
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# A manufacturer makes umbrellas at the cost of c dollars pe

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