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A certain company plans to sell Product X for p dollars per

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A certain company plans to sell Product X for p dollars per [#permalink]

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New post Updated on: 27 May 2013, 05:45
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A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?

A. 0
B. 1/100
C. 1/25
D. 99/100
E. 1

My ANS :
E-1,
Explanation –
Profit = Revenue -cost
= p(6-p)-(12-p)
= -(p^2-7p+12)
= -(p-3)(p-4)
There will be no profit for 1 =< p =< 100
So probability = 1.
Hence E


But the ans in given in the solution list is 99/100

Please explain which one is correct .

Originally posted by verycoolguy33 on 23 Feb 2012, 21:24.
Last edited by Bunuel on 27 May 2013, 05:45, edited 2 times in total.
Edited the question and added the OA
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 24 Feb 2012, 02:09
3
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A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?
A. 0
B. 1/100
C. 1/25
D. 99/100
E. 1

Profit=Revenue-Cost=p(6 - p)-(12-p).

Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6 - p)-(12-p)>0\) --> \(6p-p^2-12+p>0\) --> \(p^2-7p+12<0\) --> \((p-4)(p-3)<0\) --> \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100).

The probability that the company will NOT see a profit on sales is \(1-\frac{1}{100}=\frac{99}{100}\).

Answer: D.

Hope it's clear.
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 25 Feb 2012, 07:11
Thanks Bubuel , I understand your explanation . But not sure whether we can suddenly shift our question from finding probablity of values for P to probablity of range for P .
I understand there has a range within which there has a profit , but if we think about possible values for P within that range there has infinite values and so it is TURE for the total possible values from 1 to 100 - in line to this . So isnt it out of the question to find out the probablity of the range in which the values of P satisfy the condition instead of the probablity for the eaxct values of P .

Thanks.
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 25 Feb 2012, 08:50
verycoolguy33 wrote:
Thanks Bubuel , I understand your explanation . But not sure whether we can suddenly shift our question from finding probablity of values for P to probablity of range for P .
I understand there has a range within which there has a profit , but if we think about possible values for P within that range there has infinite values and so it is TURE for the total possible values from 1 to 100 - in line to this . So isnt it out of the question to find out the probablity of the range in which the values of P satisfy the condition instead of the probablity for the eaxct values of P .

Thanks.


I'm not sure I understand your reasoning above. Anyway there was no "sudden shift". We are asked to find the probability, not exact value in dollars, and that's exactly what we did. Consider this: what is the probability that computer will generate a random number from 3 to 4 if it's restricted to generate only numbers from 0 to 100. Favorable outcome is 1 unit out of 100, so P=1/100.

There is a little assumption though concerning the value of p, which should be some number in dollars but it has nothing to do with your doubt.
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 26 Feb 2012, 10:36
Quote:
(p-4)(p-3)<0 --> 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).


Hi buneuel,

Can you throw some light on solving efficietly equations like these

(p-4) (p-3) < 0

Though im able to pick values and justify these but in a times scenario that will not work out

also we have 3<p<4

a number less than 4 and greater than 3 can only be decimals say 3.45,3.55,3.65?
how do we coem to conclusion that it is (1/100)

please advice.
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 26 Feb 2012, 11:00
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shankar245 wrote:
Quote:
(p-4)(p-3)<0 --> 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).


Hi buneuel,

Can you throw some light on solving efficietly equations like these

(p-4) (p-3) < 0

Though im able to pick values and justify these but in a times scenario that will not work out

also we have 3<p<4

a number less than 4 and greater than 3 can only be decimals say 3.45,3.55,3.65?
how do we coem to conclusion that it is (1/100)

please advice.


Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

As for your another question: p can be decimal. Why not? The distance of a range from 3 to 4 is 1 unit, there are total of 100 units, the probability that p is in this range is favorable/total=1/100.

Hope it helps.
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 23 Jun 2014, 02:31
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?
Quote:
Profit=Revenue-Cost=p(6 - p)-(12-p).

Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: p(6 - p)-(12-p)>0 --> 6p-p^2-12+p>0 --> p^2-7p+12<0 --> (p-4)(p-3)<0 --> 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).

The probability that the company will NOT see a profit on sales is 1-\frac{1}{100}=\frac{99}{100}.

Answer: D.

Hope it's clear.


Dear Bunuel,
For the revenue to be realized the value of p can not be greater than 6, as negative revenue is not possible. So shouldn't the answer to the question be 1/6. :?:
Thanks !
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A certain company plans to sell Product X for p dollars per [#permalink]

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New post 25 Jul 2015, 22:59
parassagi wrote:
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?
Quote:
Profit=Revenue-Cost=p(6 - p)-(12-p).

Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: p(6 - p)-(12-p)>0 --> 6p-p^2-12+p>0 --> p^2-7p+12<0 --> (p-4)(p-3)<0 --> 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).

The probability that the company will NOT see a profit on sales is 1-\frac{1}{100}=\frac{99}{100}.

Answer: D.

Hope it's clear.


Hi Bunuel, I have a query on the above post. I think the sample space or possible values of p need to be from 0 to 6 only as beyond 6, 6-p becomes negative and therefore revenue becomes negative, which is not possible. So the correct answer needs to be 5/6 rather than 99/100 - what do you think? Please clarify and do correct me if I am wrong, thanks.
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A certain company plans to sell Product X for p dollars per [#permalink]

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New post 05 Apr 2017, 01:24
Bunuel wrote:
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?
A. 0
B. 1/100
C. 1/25
D. 99/100
E. 1

Profit=Revenue-Cost=p(6 - p)-(12-p).

Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6 - p)-(12-p)>0\) --> \(6p-p^2-12+p>0\) --> \(p^2-7p+12<0\) --> \((p-4)(p-3)<0\) --> \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100).

The probability that the company will NOT see a profit on sales is \(1-\frac{1}{100}=\frac{99}{100}\).

Answer: D.

Hope it's clear.


Dear Bunuel, Why the probability that company will see a profit on sales is \(\frac{1}{100}\) since \(3 < p < 4\) and p is neither 3 nor 4? :?
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Re: A certain company plans to sell Product X for p dollars per [#permalink]

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New post 05 Apr 2017, 02:00
ziyuen wrote:
Bunuel wrote:
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12 - p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6 - p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?
A. 0
B. 1/100
C. 1/25
D. 99/100
E. 1

Profit=Revenue-Cost=p(6 - p)-(12-p).

Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6 - p)-(12-p)>0\) --> \(6p-p^2-12+p>0\) --> \(p^2-7p+12<0\) --> \((p-4)(p-3)<0\) --> \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100).

The probability that the company will NOT see a profit on sales is \(1-\frac{1}{100}=\frac{99}{100}\).

Answer: D.

Hope it's clear.


Dear Bunuel, Why the probability that company will see a profit on sales is \(\frac{1}{100}\) since \(3 < p < 4\) and p is neither 3 nor 4? :?


Are you asking why is the probability 1/100 even though it's \(3 < p < 4\) and not \(3 \leq p \leq 4\)? If so, then 3 and 4 are points with no dimension, thus the probaiblity that p is in \(3 < p < 4\) is the same as the probability that p is in \(3 \leq p \leq 4\).
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Re: A certain company plans to sell Product X for p dollars per   [#permalink] 05 Apr 2017, 02:00
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