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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?

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

Answer is C
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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 Answer is C

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

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

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 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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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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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?

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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