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If x and y are consecutive odd integers such that x < y

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If x and y are consecutive odd integers such that x < y [#permalink]

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New post 19 Sep 2012, 13:38
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If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.


See the answer and a full video explanation here.
http://gmat.magoosh.com/questions/880
[Reveal] Spoiler: OA

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Re: If x and y are consecutive odd integers such that x < y [#permalink]

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New post 19 Sep 2012, 16:09
If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative. x and y have the opposite signs. Consecutive odd integers to have the opposite signs one must be equal to -1 and another to 1, so their sum is -1+1=0. Sufficient.

(2) The sum x + y is the square of an integer. If x=-1 and y=1, then -1+1=0^2 but if x=1 and y=3, then x+y=2^2. Not sufficient.

Answer: a.
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Re: If x and y are consecutive odd integers such that x < y [#permalink]

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New post 19 Sep 2012, 20:48
mikemcgarry wrote:
If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.


See the answer and a full video explanation here.
http://gmat.magoosh.com/questions/880


(1) xy is negative only if one number is negative and the other is positive. (-1) and (1) is the only possibility. Thus. SUFFICENT
(2) x + y = s^2

Try (-1) + (1) = 0, is equal to 0^2
Try (1) + (3) = 4, is equal to 2^2, Thus, INSUFFICIENT

Answer: A
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Re: If x and y are consecutive odd integers such that x < y [#permalink]

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New post 20 Sep 2012, 00:11
mikemcgarry wrote:
If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.


See the answer and a full video explanation here.
http://gmat.magoosh.com/questions/880


(1) Obviously one integer must be negative and the other one positive. Being both odd, the only possibility is -1 and 1.
Sufficient.

(2) There are infinitely many pairs of consecutive integers for which their sum is a square of an integer.
If \(x\) and \(y\) are consecutive odd integers, \(x<y\), then between them there is a unique even integer \(2k\), where \(2k=(x+y)/2.\)
\(x=2k-1\) and \(y=2k+1.\) Their sum is \(4k\) and necessarily \(4k=n^2\), or \(2k=n^2/2.\)

So, take any even integer \(n\), then \(\frac{n^2}{2}-1\) and \(\frac{n^2}{2}+1\) are two consecutive odd integers.
Tor example, 1 and 3, 7 and 9, 17 and 19, 31 and 33,...

Not sufficient.

Answer A
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Re: If x and y are consecutive odd integers such that x < y [#permalink]

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New post 16 May 2013, 11:31
mikemcgarry wrote:
If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.


See the answer and a full video explanation here.
http://gmat.magoosh.com/questions/880


Consecutive odd integers and product is negative => -1 and 1 :)

stmt 2: nope wrong, can't get any particular value.

so A

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Re: If x and y are consecutive odd integers such that x < y [#permalink]

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New post 03 Oct 2017, 14:39
If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.
Just by picking numbers: -1*-3 =3 ; 1*3 = 3. Only number that does this is -1*1 (with zero in the middle) is negative. so SUFF.

(2) The sum x + y is the square of an integer.-3+(-1) = -4. But -4 cannot be square of any integer. 3+1 = is sq of 2. 1 + (-1)
= 0; which is not square of anything. So different answers.


Answer is A

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Re: If x and y are consecutive odd integers such that x < y   [#permalink] 03 Oct 2017, 14:39
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