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.

QUICK SOLUTION FOR THIS PROBLEM

(1) Clearly insufficient.

(2) From the statement, we know that:
C< 1.2A
C<1.2B
C<1.2D
...
C<1.2I

If we sum all of these inequations, we get:
9.C < 1.2 . (21000 - C)

Therefore,
10.2 C < 25200
C< aprox 2471 < 3000 (SUFFICIENT)

ANS B
_________________

Please press "kudo" if this helped you!

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.
_________________

In the study cave!

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.