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If p and q are consecutive even integers, is p>q? 1) p-2

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If p and q are consecutive even integers, is p>q? 1) p-2 [#permalink] New post 06 Jun 2007, 09:59
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If p and q are consecutive even integers, is p>q?

1) p-2 and q+2 are consecutive even integers
2) p is prime
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 [#permalink] New post 06 Jun 2007, 10:31
a) True only for p </= 0 AND q <0> p-2=-2 & q+2=0 (Cons. Even) implies p>q for all [p, q].SUFF


b) P=PRIME and EVEN=2 implies q = 4. i.e. p<q...............SUFF.
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 [#permalink] New post 06 Jun 2007, 10:39
I think it is B. I shall explain further if I am right
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 [#permalink] New post 06 Jun 2007, 13:55
go with (A)

given p and q are consecutive even integers.

St1 : (p-2) and (q+2) are consecutive even integers

now if we assume p>q,

p-2 = q + 2 +2 => p = q + 6 ; but if p = q +6, p and q cannot be consecutive even

if we assume p<q> p = q+2 --- SUFF


St2 : p is prime => p = 2 , q can be either '0' or '4' INSUFF

Hence (A)
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Re: DS - consecutive even [#permalink] New post 06 Jun 2007, 14:22
Caas wrote:
If p and q are consecutive even integers, is p>q?

1) p-2 and q+2 are consecutive even integers
2) p is prime


(1) if p-2 and q+2 are consecutive also, then it MUST be the case that p>q. Suff.

(2) p is prime, then it´s =2. If p and q are consecutive even integers, then q = 0 or 4 (even integer can be +ve, -ve, or 0), then we cannot know for sure whether p<>q. Insuff.

A.
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Re: DS - consecutive even [#permalink] New post 06 Jun 2007, 15:05
I'll go with A.

1) P-2 and q+2 are consecutive even integers.
take the sample numbers where both numbers are positive and both are negative. you will see that p>q is always true for p-2 and q +2 to be consecutive even integers. you can also test this eith {-2, 0} and {0, 2}. - sufficient

2) p is prime. q could be 0 or 4. for 0 p >q holds up but not for 4. - not sufficient

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 [#permalink] New post 07 Jun 2007, 11:38
Hey guys,
when i first looked at it, it also thought it was A, but if yo uthink about it's not enough to just say that p IS greater the q. In this case q is greater than p. (from the second option, where p is prime) The question asks if p>q, in this case you could you with certainty that it is not, therefore D, wither statement is suffiecient. One proves that p>q, and the other that P<q. Who cares if it is or not, point is that you could say it with certinty, and this is what GMAT wants.
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 [#permalink] New post 07 Jun 2007, 12:30
Stmnt1:

p=6 q=4

6-2=4, 4+2=6 in this case p is greater than q. 6>4. You gotta take what you plugged in for p and q. Not the result of the subtraction.



B is insuff. While P has to be 2.

Q could be 0 or 4 since 0 is considered an even integer.


Answer A
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 [#permalink] New post 07 Jun 2007, 12:32
For this DS question, I will examine each statement alone.

What's asked: is p>q ?
What's given: p and q are consecutive even integers

What do we know?
either p = q+2 or p = q-2

1) p-2 and q+2 are consecutive even integers
-------------------------------------------------------
if p = q-2, then p-2 = q-4 which can never be a consecutive even integer after or before q+2
Thus, p = q+2 and hence p>q

Statement 1 is sufficient

2) p is prime
----------------
We know that p is even, and the only even prime integer is 2. Since q and p are consecutive even integers, q can be either 0 or 4.

Statement 2 is insufficient


My Answer: A
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 [#permalink] New post 07 Jun 2007, 21:37
OA is A :-D
Thank you all!
  [#permalink] 07 Jun 2007, 21:37
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