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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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emmak
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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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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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 - https://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - https://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - https://www.gmatprepnow.com/module/gmat- ... /video/935

Cheers,
Brent
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emmak
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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emmak
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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emmak
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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emmak
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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emmak
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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Maybe I complicated things but I used fractions of x to get the equation:

2cx = rx(1-y) + bxy

2cx = rx - rxy + bxy

2cx - rx = xy(b-r)

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

xy = the fraction of the total number of umbrellas that can be sold for b dollars.
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emmak
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)


Say, he sells N umbrellas at below cost rate of b and rest of (x - N) umbrellas at retail price of r dollars.
how do we know this?
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