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Manager  Joined: 09 Feb 2013
Posts: 114
A manufacturer makes umbrellas at the cost of c dollars pe  [#permalink]

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Question Stats: 60% (02:53) correct 40% (03:10) wrong based on 210 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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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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Intern  Joined: 08 Feb 2013
Posts: 3
Re: A manufacturer makes umbrellas at the cost of c dollars pe  [#permalink]

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8
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
SVP  Joined: 06 Sep 2013
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Concentration: Finance
Re: A manufacturer makes umbrellas at the cost of c dollars pe  [#permalink]

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

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

Thanks!
Cheers!
J Manager  Joined: 02 Apr 2012
Posts: 67
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34 WE: Consulting (Consulting)
Re: A manufacturer makes umbrellas at the cost of c dollars pe  [#permalink]

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

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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 c = 2, x = 10, r = 5 and b = 0 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 _________________ Originally posted by GMATPrepNow on 22 Oct 2015, 11:39. Last edited by GMATPrepNow on 26 Feb 2019, 07:13, edited 1 time in total. CEO  V Joined: 12 Sep 2015 Posts: 3726 Location: Canada Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink] ### Show Tags 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 _________________ Senior SC Moderator V Joined: 14 Nov 2016 Posts: 1334 Location: Malaysia A manufacturer makes umbrellas at the cost of c dollars pe [#permalink] ### Show Tags 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. _________________ "Be challenged at EVERY MOMENT." “Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.” "Each stage of the journey is crucial to attaining new heights of knowledge." Rules for posting in verbal forum | Please DO NOT post short answer in your post! Advanced Search : https://gmatclub.com/forum/advanced-search/ Veritas Prep GMAT Instructor D Joined: 16 Oct 2010 Posts: 9238 Location: Pune, India Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink] ### Show Tags 1 1 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) _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Manager  G Joined: 07 Aug 2018 Posts: 110 Location: United States (MA) GMAT 1: 560 Q39 V28 GMAT 2: 670 Q48 V34 Re: A manufacturer makes umbrellas at the cost of c dollars pe [#permalink] ### Show Tags 1 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. --> $$2cx= N*b + (x-N)*r$$ -->$$N = \frac{(2cx - rx)}{(b - r)}$$ _________________ Senior Manager  G Joined: 04 Aug 2010 Posts: 415 Schools: Dartmouth College A manufacturer makes umbrellas at the cost of c dollars pe [#permalink] ### Show Tags 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. $$\frac{2(xc-r)}{(x-r)}$$ 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 -- 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.

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

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

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

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