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Re: M15#11 [#permalink]
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Samwong wrote:
Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0

If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5

(2) xy = 6


Actually you have more than one constraint:
(i) x and y are integers;
(ii) x<y<0

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Answer: A.

Hope it helps.
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Re: M15#11 [#permalink]
Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?
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Re: M15#11 [#permalink]
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pavanpuneet wrote:
Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?


Yes, since both sides of the inequality are non-negative then we can square it and write: (x-y)^2>(x+y)^2.

Generally:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Hope it helps.
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Re: M15#11 [#permalink]
For the life of me I can't see a situation where abs (x+y) is less than 2, which would make statement B sufficient.

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