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Which number in set P has a value greater than that of every

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Which number in set P has a value greater than that of every [#permalink] New post 07 Sep 2005, 05:22
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Which number in set P has a value greater than that of every other member of all set P, if all the members of set P are negative integers?

1 Each member of set P is the product of -1 and a prime number
2At least one member of P is even
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 [#permalink] New post 07 Sep 2005, 08:25
I'd go with C.

If -2 is a member of the set, then it would be the greatest number. But without statement 2, we don't know that -2 is a member of the set. In fact, from the stem we don't know how many numbers are in set P at all. Perhaps set P={-7,-11,-17}.

Since there's only one product of -1 and a prime number that can be even, and statement 2 tells us it's in the set, then both in combination are sufficient. So C.
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 [#permalink] New post 07 Sep 2005, 08:43
A for me
statement 1 is suff -> we know all the numbers in set P are negative integers so the biggest would be -1. Then we know that each one is a oproduct if -1 and a prime number so the biggest has to be -2

statement 2 gives us nothing.
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 [#permalink] New post 07 Sep 2005, 08:51
think it is E)
from A) it can be any prime so insuff
from B) it is also insuff
taken both, the only even prime is 2 , but the Q says "at least" which means that there might be more than one 2-s so the ans seems E)
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 [#permalink] New post 07 Sep 2005, 11:40
I go for C here.

Stmt one tells us that each member of the set is a negative prime but it doesn't say how many primes or which one it starts with. The set could start at -3 or -13 or only have -3 and -13 in it. Insufficient

Stmt two is obviously insufficient.

Put together and sufficient. We can answer the question since there is only one even prime number.
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Re: DS:Princeton Rev 2° [#permalink] New post 07 Sep 2005, 18:17
Macedon wrote:
Which number in set P has a value greater than that of every other member of all set P, if all the members of set P are negative integers?

1 Each member of set P is the product of -1 and a prime number
2At least one member of P is even


This question isn't very clear. If it is asking what is the maximum value in P then 1 and 2 together would be sufficient to determine that -2 is in P. Since -2 is greater than all the other products of -1 and prime numbers, then we can say the answer is C.

However if it is asking which element has the greatest value, then the answer would be E, since there could be more than one element that has the value of -2.
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Re: DS:Princeton Rev 2°   [#permalink] 07 Sep 2005, 18:17
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