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John owns stock in nine different companies, for a total of [#permalink]

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24 Mar 2012, 09:58

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John owns stock in nine different companies, for a total of 21000 shares. If the number of shares he has in any company is not equal to the number of shares he has in any other company, does he own at least 3000 shares of C?

(1) John has more shares of company C stock than any other company's stock, except for company A and B.

(2) The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns.

John owns stock in nine different companies, for a total of 21,000 of stock. If the number of shares he has in any company is not equal to the number of shares he has in any other company, does he own at least 3,000 shares of Command C stock? 1) John has more shares of Company C stock than any other company's stock, except for Company A and Company B. 2) The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns.

This is a hard question. I am happy to help with this.

So, John owns share from nine companies, for a total of 21000 shares. Number of shares from each company unequal to all the others.

Statement #1: John has more shares of Company C stock than any other company's stock, except for Company A and Company B. We could construct scenarios in which the answer to the prompt is true or false For example A = 20000 shares, B = 979 shares, C = 7, D = 6, E = 5, F = 4, G = 3, H = 2, I = 1 This extreme scenario gives an answer of no. A = 5500, B = 4500, C = 4000, D= 3000, E = 2500, F = 2000, G = 1500, H = 1000, I = 500 This scenario give an answer of yes. Because we can construct scenarios which answer the prompt question either way, we have no way to give a definitive answer based on this sentence. Therefore, this sentence, by itself, is insufficient.

Statement #2: The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns Clearly, in an extreme case, John could own 1 share of C, and more than 1 share of everything else, and the answer would be no. The question, then, remains, would it be possible to construct a scenario consistent with this statement that give a "yes' answer to the prompt?

In order for C to be as big as possible, we would need the minimum to be as large as possible, which implies that the values on the list are all close together.

If C = 3000, that leaves 18000 for the other eight companies, which is an average of 18000/8 = 9000/4 = 2250 shares each. Of course, they couldn't all equal that value, so some would have to be slightly above that, and others, including the minimum, would have to be slightly below that.

What is 20% more than 2250? 1.20*2250 = 2700, far less than 3000.

Thus, it turns out, given the constraint of this statement, it's absolutely impossible for C to be as large as 3000. It absolutely must be smaller than 3000. This statement is sufficient.

Re: John owns stock in nine different companies, for a total of [#permalink]

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11 Nov 2012, 12:58

John owns stock in nine different companies, for a total of 21,000 of stock. If the number of shares he has in any company is not equal to the number of shares he has in any other company, does he own at least 3,000 shares of Command C stock? 1) John has more shares of Company C stock than any other company's stock, except for Company A and Company B. 2) The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns.
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Re: John owns stock in nine different companies, for a total of [#permalink]

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11 Nov 2012, 18:47

BN1989 wrote:

John owns stock in nine different companies, for a total of 21000 shares. If the number of shares he has in any company is not equal to the number of shares he has in any other company, does he own at least 3000 shares of C?

(1) John has more shares of company C stock than any other company's stock, except for company A and B.

(2) The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns.

Assume that C stock is 12% and the rest 8 companies 11% total 100%,

No. of share of C 12 % of 21000 < 3000

If C is around 15 % and we get 15% of 21000 > 3000

Re: John owns stock in nine different companies, for a total of [#permalink]

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10 Apr 2013, 15:25

karthikguptav wrote:

BN1989 wrote:

John owns stock in nine different companies, for a total of 21000 shares. If the number of shares he has in any company is not equal to the number of shares he has in any other company, does he own at least 3000 shares of C?

(1) John has more shares of company C stock than any other company's stock, except for company A and B.

(2) The number of shares of company C stock that John owns is less than 20% greater than the number of shares of any other company's stock that John owns.

Assume that C stock is 12% and the rest 8 companies 11% total 100%,

No. of share of C 12 % of 21000 < 3000

If C is around 15 % and we get 15% of 21000 > 3000

Even B is not sufficient

Remember that A and B are both more shares than C.

Let's assume that every number is AS LOW AS POSSIBLE to give C the most possible shares, since if we are definitively able to say that C is either less than or greater than 3000, it'll probably be less than based off of 21000/9 < 3000.

If C = 3000, and we use the informatino in part I, then B is at minimum 3001 and A is at minimum 3002. (again, we're using minimums to give C the most value possible to reach over 3000)

Therefore, if C is 3000 or greater, the remaining 6 have to be equal to 11,997 or LESS (let's call it 12000).

With B, if we assume that ALL of the remaining choices are within 20% of C, and we want to assume that they're as low as possible in order to give C the highest chance of being 3k or over, then the least they could be is = (3000 / 1.2).

Because there are SIX of them, and all of them have to be at least 3000 / 1.2, we get that the sum of the six has to be at least (3000) / (1.2) * (6)... the six and the 1.2 factors out neatly into a five, which means that the sum of the remaining six is AT LEAST 15000.

If we put the two parts together, in order for C to be 3000 or greater than the sum of the remaining six has to be both over 15000 and under 12000. This is impossible, so part 1 and part 2 of the information, combined, let us say definitively that the answer is no.

If we just use part II of the information, however, we get that, in order for A to be 3000 or greater, the sum of the remaining EIGHT choices must equal 2700 or less.

Combine this with the "less than 20% greater" and we have a lot of possibilities that do not definitively tell us anything.

Answer must be C.

EDIT: whoops i'm wrong. If C is 3000 or more, then the remaining 8 types of shares is = 18000 or less.

Using part II only, if C is 3000 or more, then the least the each of them can be is 3000 * .8 = 2400 or more. If there are 8 of them, then the sum of the 8 of them must be 19200 or more. Therefore, in order for C to be more than 3000, the sum of the remaining shares must be both less than 18000 and more than 19200... which is impossible.

Re: John owns stock in nine different companies, for a total of [#permalink]

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12 May 2013, 07:00

Quiet similar to what othrs have done ----

Stmt 1 : clearly insufficient Stmt 2 : we can pick a very low value for C such that c<3000

Now to get the highest value for C - it is given that c< 1.2 (A|B|D|E|F|G) i.e ( A or B or D or E or F or G)

now lets say A is the highest and G is the lowest among A or B or D or E or F or G

Case 1 : You get a VERY LOW value for C if :- if u make A too high then G would be too low ( since sum has to total to 21000 , including C) . And C would be lower than 1.2G ( as per Stmt 2) . Hence you will get a low value for C Case 2: As per above logic , You get a HIGHEST value for C if :- , you minimize the difference among the remaining 6 companies . Lets say they all average to A

Hence C + 1.2 ( 6A) = 21000 ( note that this is the highest value for C). If C is equal to A , then we would get - 8.2A =21000 Hence A = 21000/8.2 ~ 2725 ( slightly less than 2725).

But C is <1.2A ( or rather C is less than A ) > Hence C has to be less than 2725 ( i.e 3000 ). Hence B is sufficient.

Re: John owns stock in nine different companies, for a total of [#permalink]

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24 Oct 2015, 07:00

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