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For Manufacturer M, the cost C of producing X units of its [#permalink]

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28 Oct 2006, 14:49

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59% (01:44) correct 41% (01:38) wrong based on 361 sessions

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For Manufacturer M, the cost C of producing X units of its product per month is given by C = kx + t, where C is in dollars and k and t are constants. Last month, if Manufacturer M produced 1,000 units of its product and sold all the units for k + 60 dollars each, what was manufacturer M's gross profit on the 1000 units?

(1) Last month, Manufacturer M's revenue from the sale of the 1000 units was $150,000 (2) Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1,000 units in a month.

Re: DS: Gross profit for Manufacturer M [#permalink]

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28 Oct 2006, 17:22

anandsebastin wrote:

For Manufacturer M, the cost C of producing x units of its product per month is given by C= kx+t, where c is in dollars and k and t are constants. Last month, if Manufacturer M produced 1,000 units of its product and sold all the units for k+60 dollars each, what was Manufacturer M's gross profit on the 1,000 units?

(1) Last month, Manufacturer M's revenue from the sale of the 1,000 units was $150,000.

(2) Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1,000 units in a month.

(OA may be late. Sorry!)

This smells like C. Profit=Revenue-Cost

1) tells us total revenue for 1,000 units. now we need cost
2) tells us that 500 units cost 45,000, so 1,000 units cost 90,000

Re: DS: Gross profit for Manufacturer M [#permalink]

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29 Oct 2006, 07:12

anandsebastin wrote:

For Manufacturer M, the cost C of producing x units of its product per month is given by C= kx+t, where c is in dollars and k and t are constants. Last month, if Manufacturer M produced 1,000 units of its product and sold all the units for k+60 dollars each, what was Manufacturer M's gross profit on the 1,000 units?

(1) Last month, Manufacturer M's revenue from the sale of the 1,000 units was $150,000.

(2) Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1,000 units in a month.

(OA may be late. Sorry!)

Looks to be B

1) 1,000k+t=1,000(k+60)-150,000
solve, and you get
t=-90,000

How do you solve this one what are the equations! What's the OA?

For Manufacturer M, the cost C of producing X units of its product per month is given by C = kx + t, where C is in dollars and k and t are constants. Last month, if Manufacturer M produced 1,000 units of its product and sold all the units for k + 60 dollars each, what was manufacturer M's gross profit on the 1000 units?

(1) Last month, Manufacturer M's revenue from the sale of the 1,000 units was $150,000 --> \(1,000(k + 60) = 150,000\) --> \(k=90\). We still don't know the value of t. Not sufficient.

(2) Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1,000 units in a month. Since given that \(C = kx + t\), then from this statement we have that \((500k+t)+45.000=1000k+t\) --> cancel t: \(500k+45.000=1000k\) --> \(k=90\). We still don't know the value of t. Not sufficient.

Re: For Manufacturer M, the cost C of producing X units of its [#permalink]

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16 Aug 2014, 03:34

Did this problem twice and still missed it. If you look closely, you will realize that the problem can be reduced to "find the value of t". Both (1) and (2) only give k, so the answer is E.

Re: For Manufacturer M, the cost C of producing X units of its [#permalink]

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14 Jan 2017, 12:52

c=kx+t In last month, cost=1000k+t; profit=1000(k+60) -(1000k+t) =60000-t, so, we need to solve t. From 1, 150000=1000*(k+60), there is no information about t. From 2, (1000k+t)-(500k+t)=45000, still cannot solve out t. Answer is E
_________________

For Manufacturer M, the cost C of producing X units of its product per month is given by C = kx + t, where C is in dollars and k and t are constants. Last month, if Manufacturer M produced 1,000 units of its product and sold all the units for k + 60 dollars each, what was manufacturer M's gross profit on the 1000 units?

(1) Last month, Manufacturer M's revenue from the sale of the 1000 units was $150,000 (2) Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1,000 units in a month.

We are given the formula c = kx + t, in which c is in dollars, x is the number of units produced per month, and k and t are constants. We are also given that last month manufacturer M produced 1000 units of its product and sold all the units for k + 60 dollars each. We need to determine manufacturer M's gross profit on the 1000 units. Thus, we have:

c = k(1,000) + t

Furthermore, since the manufacturer sells 1000 units for k + 60 dollars each, its revenue was 1000(k + 60) = 1000k + 60000. So, the gross profit is 1000k + 60000 - (1000k + t ) = 60000 - t. As we can see, if we can determine the value of t, then we can answer the question.

Statement One Alone:

Last month, manufacturer M's revenue from sale of the 1000 units was $150,000.

Using the information in statement one, we can create the following equation:

1000k + 60,000 = 150,000

1000k = 90,000

k = 90

Although we have a value for k, we have no information about the value of t. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

Manufacturer M's cost of producing 500 units in a month is $45,000 less than its cost of producing 1000 units in a month.

Using the information in statement two, we can create the following equation:

k(1,000) + t - [k(500) + t] = 45,000

1,000k + t - 500k - t = 45,000

500k = 45,000

k = 90

Although we have a value for k, we still do not have enough information to determine the value of t. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Since each statement allows us to find only the value of k (recall k = 90 in each statement), we still do not have enough information to determine the value of t, so we cannot determine the profit last month.

Answer: E
_________________

Jeffery Miller Head of GMAT Instruction

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