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

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.

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.

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?


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 !

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.

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 \(3 \leq p \leq 4\).

how do we find out the level of difficulty (600 level or 700 level) of this question?

thanks

The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. So our stats say that it's 700 level.

What if the options have 98/100. having picked the range 3 and 4, we know that the company will make a profit at and inside these values, so 1- 2/100=98/100. what are your views on this?

Why isn't option C the answer.

I substituted the value for p in both manufacturing cost and revenue. Took the range for value p as 12<=p<0(as p should be a positive number and 0 is not a positive number).Got 4 values for which the manufacturing cost > revenue(hence loss). ie when p = 1,2,4,5.So shouldn't the answer be 4/100 or 1/25?

Please advice.

Given: 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).

Asked: 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?

Selling price = p dollars/unit
p<100
Monthly Revenues = 1000p(6-p) dollars
Monthly Manufacturing cost = 1000(12-p) dollars
Units sold = 1000(6-p)
Profits = Revenues - Costs = 1000p(6-p) - 1000(12-p) = 1000(6p - p^2-12+p) = 1000(7p - p^2 - 12) = - 1000(p-3)(p-4)
Profits < 0 if p>4 or p<3
Profits > 0 if 3<p<4
Since 0<p<100

Probability that the company will NOT see a profit on sales of Product X in the first month of sales = 99/100

IMO D

Between 3.1 and 3.99 there are 99 units not 100. P does not include value of extremities 3 and 4. Can you please help me understand this

Where do 3.1 and 3.99 come from? Why 3.1? Why not 3.0000001 or 3.000000000000000000000000001? Anyway, it's as simple as this, the distance between 3 and 4 is 1 unit and the distance between 0 and 100 is 100 units, so P = 1/100.

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