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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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New post Updated on: 27 Feb 2013, 05:53
5
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

60% (02:49) correct 40% (02:36) wrong based on 256 sessions

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

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Originally posted by emmak on 27 Feb 2013, 04:16.
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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New post 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 = \(\frac{x(2c-r)}{(b-r)}\)

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

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New post 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
Answer is C
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Re: A manufacturer makes umbrellas at the cost of c dollars pe  [#permalink]

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

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New post 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
Answer is C

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

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New post 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

Answer : C

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

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

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New post 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.

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

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New post 25 Sep 2018, 22:44
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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. \(\frac{2(xc-r)}{(x-r)}\)


Responding to a pm:

Cost of x umbrellas = c*x
Profit he intends to make is 100% i.e. = c*x
So revenue must be = 2c*x

Say, he sells N umbrellas at below cost rate of b and rest of (x - N) umbrellas at retail price of r dollars.

\(2c*x = N*b + (x - N)*r\)

\(N = \frac{(2cx - rx)}{(b - r)}\)

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

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New post 27 Sep 2018, 01:06
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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. \(\frac{2(xc-r)}{(x-r)}\)


Had a very hard time understanding the above solutions...
So I tried to put it into simpler terms:

\(Profit = Revenue-Costs\)

\(Costs: c*x\)

\(Profit = Revenue - cx\)

Question stem tells us that the manufacturer needs to make a 100% profit. How do we get a 100% profit? Only when the profit is equal to the costs and the revenue is equal to 2 times the costs.

\(cx = Revenue - cx\)
\(cx = 2cx - cx\)

Therefore the revenue the manufacturer needs to make is equal to \(2cx\)! Now setting up the real equation:


Revenue (In terms of cost) = Revenue (In terms of below cost rate and normal cost rate)

-->\(2cx=\) Revenue below costs + Revenue normal costs

Lets use N for the #of terms that sell below cost price.

--> \(2cx= N*b + (x-N)*r\)

-->\(N = \frac{(2cx - rx)}{(b - r)}\)
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