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Is x y < x + y?
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08 Aug 2012, 22:08
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Is x – y < x + y? (1) y < x (2) xy < 0
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Re: Is x y < x + y?
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08 Aug 2012, 23:21
1. y<x implies 2 cases: (a) y is +ve, x is +ve. eg., y=2, x=3 xy = 1 x = 3 y = 2 x + y = 5, Hence xy < x + y
(b) y is ve, x is +ve. eg., y=1, x = 2 xy=3 x+y=3 Here, xy !< x+y.
From (a) and (b), A and D are eliminated
2. xy<0 <=> either x or y is ve. Hence xy is always = x + y So, here we have a definite answer to the question and hence B is the best choice.



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Re: Is x y < x + y?
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09 Aug 2012, 00:57
Is x – y < x + y? Since both sides of the inequality are nonnegative, then we can safely square and we'll get: is \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? Consider two cases: If \(xy\leq{0}\), then \(xy+xy=xy+xy=0\). So, when \(xy\leq{0}\), then \(xy+xy>0\) does NOT hold true. If \(xy>0\), then \(xy+xy=xy+xy=2xy>0\) (because \(xy>0\)). So, when \(xy>{0}\), then \(xy+xy>0\) holds true. As we can see the question basically asks whether \(xy>0\). (1) y < x. Not sufficient to say whether \(xy>0\). (2) xy < 0. Directly gives a NO answer to the question. Answer: B. Hope it's clear.
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Re: Is x y < x + y?
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09 Aug 2012, 04:55
venmic wrote: Is x – y < x + y?
(1) y < x
(2) xy < 0
Bunuel
I searched of this question on gmatclub culd not find a post please explain your method Useful to remember the socalled "Triangle Inequality " for the absolute value: For any real numbers \(x,y\) \(\,x+y\leqx+y\). Equality holds if and only if \(xy\geq{0}\), which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative). Or, strict inequality holds if and only if \(xy<0.\) In our case, we can see that always \(xy=x+(y)\leqx+y\), because \(y=y\). The strict inequality will hold if and only if \(x(y)<0\) or \(xy>0.\) (1) Either \(x\) or \(y\) can be zero, and in addition, we don't know anything about their signs. Not sufficient. (2) Obviously sufficient, because we can state with certainty that the given inequality does not hold. Answer B
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Re: Is x y < x + y?
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09 Aug 2012, 05:00
EvaJager wrote: venmic wrote: Is x – y < x + y?
(1) y < x
(2) xy < 0
Bunuel
I searched of this question on gmatclub culd not find a post please explain your method Useful to remember the socalled "Triangle Inequality " for the absolute value: For any real numbers \(x,y\) \(\,x+y\leqx+y\). Equality holds if and only if \(xy\geq{0}\), which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative). Or, strict inequality holds if and only if \(xy<0.\) In our case, we can see that always \(xy=x+(y)\leqx+y\), because \(y=y\). The strict inequality will hold if and only if \(x(y)<0\) or \(xy>0.\) (1) Either \(x\) or \(y\) can be zero, and in addition, we don't know anything about their signs. Not sufficient. (2) Obviously sufficient, because we can state with certainty that the given inequality does not hold. Answer B There is another property worth remembering: 1. Always true: \(x+y\leq{x+y}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign); 2. Always true: \(xy\geq{xy}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(x>y\) (simultaneously).
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Re: Is x y < x + y?
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09 Aug 2012, 05:08
Bunuel wrote: EvaJager wrote: venmic wrote: Is x – y < x + y?
(1) y < x
(2) xy < 0
Bunuel
I searched of this question on gmatclub culd not find a post please explain your method Useful to remember the socalled "Triangle Inequality " for the absolute value: For any real numbers \(x,y\) \(\,x+y\leqx+y\). Equality holds if and only if \(xy\geq{0}\), which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative). Or, strict inequality holds if and only if \(xy<0.\) In our case, we can see that always \(xy=x+(y)\leqx+y\), because \(y=y\). The strict inequality will hold if and only if \(x(y)<0\) or \(xy>0.\) (1) Either \(x\) or \(y\) can be zero, and in addition, we don't know anything about their signs. Not sufficient. (2) Obviously sufficient, because we can state with certainty that the given inequality does not hold. Answer B There is another property worth remembering: 1. Always true: \(x+y\leq{x+y}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign); 2. Always true: \(xy\geq{xy}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(x>y\) (simultaneously). Thanks for reminding, certainly worth remembering. Just for the practice, we can prove that the second inequality can be deduced from the first one quite easily: \((xy)+y\leqxy+y\), or \(x\leqxy+y\), from which we get \(xy\geq{xy}.\)
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Re: Is x y < x + y?
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06 Jun 2013, 03:56
Ans B Ref to the attached pic Stmt 1: has two possibilities Stmt 2 : gives confirmed no Therefore ans B
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Re: Is x y < x + y?
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10 Jun 2013, 11:29
Hi. I am wondering how you got: \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with 2xy while on the right side you are left with 2xy Thanks! Bunuel wrote: Is x – y < x + y?
Since both sides of the inequality are nonnegative, then we can safely square and we'll get: is \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? Consider two cases:
If \(xy\leq{0}\), then \(xy+xy=xy+xy=0\). So, when \(xy\leq{0}\), then \(xy+xy>0\) does NOT hold true. If \(xy>0\), then \(xy+xy=xy+xy=2xy>0\) (because \(xy>0\)). So, when \(xy>{0}\), then \(xy+xy>0\) holds true.
As we can see the question basically asks whether \(xy>0\).
(1) y < x. Not sufficient to say whether \(xy>0\).
(2) xy < 0. Directly gives a NO answer to the question.
Answer: B.
Hope it's clear.



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Re: Is x y < x + y?
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10 Jun 2013, 11:57
WholeLottaLove wrote: Hi. I am wondering how you got: \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with 2xy while on the right side you are left with 2xy Thanks! Bunuel wrote: Is x – y < x + y?
Since both sides of the inequality are nonnegative, then we can safely square and we'll get: is \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? Consider two cases:
If \(xy\leq{0}\), then \(xy+xy=xy+xy=0\). So, when \(xy\leq{0}\), then \(xy+xy>0\) does NOT hold true. If \(xy>0\), then \(xy+xy=xy+xy=2xy>0\) (because \(xy>0\)). So, when \(xy>{0}\), then \(xy+xy>0\) holds true.
As we can see the question basically asks whether \(xy>0\).
(1) y < x. Not sufficient to say whether \(xy>0\).
(2) xy < 0. Directly gives a NO answer to the question.
Answer: B.
Hope it's clear. \(x^22xy+y^2<x^2+2xy+y^2\) > cancel x^2+y^2 in both sides: \(2xy<2xy\) > reduce by 2: \(xy<xy\) > \(xy+xy>0\). Hope it's clear.
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Re: Is x y < x + y?
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10 Jun 2013, 12:04
I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't. Thanks! Bunuel wrote: WholeLottaLove wrote: Hi. I am wondering how you got: \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with 2xy while on the right side you are left with 2xy Thanks! Bunuel wrote: Is x – y < x + y?
Since both sides of the inequality are nonnegative, then we can safely square and we'll get: is \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? Consider two cases:
If \(xy\leq{0}\), then \(xy+xy=xy+xy=0\). So, when \(xy\leq{0}\), then \(xy+xy>0\) does NOT hold true. If \(xy>0\), then \(xy+xy=xy+xy=2xy>0\) (because \(xy>0\)). So, when \(xy>{0}\), then \(xy+xy>0\) holds true.
As we can see the question basically asks whether \(xy>0\).
(1) y < x. Not sufficient to say whether \(xy>0\).
(2) xy < 0. Directly gives a NO answer to the question.
Answer: B.
Hope it's clear. \(x^22xy+y^2<x^2+2xy+y^2\) > cancel x^2+y^2 in both sides: \(2xy<2xy\) > reduce by 2: \(xy<xy\) > \(xy+xy>0\). Hope it's clear.



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Re: Is x y < x + y?
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10 Jun 2013, 12:05
I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't. Thanks! Bunuel wrote: WholeLottaLove wrote: Hi. I am wondering how you got: \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with 2xy while on the right side you are left with 2xy Thanks! Bunuel wrote: Is x – y < x + y?
Since both sides of the inequality are nonnegative, then we can safely square and we'll get: is \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)? Consider two cases:
If \(xy\leq{0}\), then \(xy+xy=xy+xy=0\). So, when \(xy\leq{0}\), then \(xy+xy>0\) does NOT hold true. If \(xy>0\), then \(xy+xy=xy+xy=2xy>0\) (because \(xy>0\)). So, when \(xy>{0}\), then \(xy+xy>0\) holds true.
As we can see the question basically asks whether \(xy>0\).
(1) y < x. Not sufficient to say whether \(xy>0\).
(2) xy < 0. Directly gives a NO answer to the question.
Answer: B.
Hope it's clear. \(x^22xy+y^2<x^2+2xy+y^2\) > cancel x^2+y^2 in both sides: \(2xy<2xy\) > reduce by 2: \(xy<xy\) > \(xy+xy>0\). Hope it's clear.



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Re: Is x y < x + y?
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10 Jun 2013, 12:14
WholeLottaLove wrote: I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't. Thanks! Bunuel wrote: WholeLottaLove wrote: Hi.
I am wondering how you got: \(x^22xy+y^2<x^2+2xy+y^2\)? > is \(xy+xy>0\)?
I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with 2xy while on the right side you are left with 2xy
Thanks!
\(x^22xy+y^2<x^2+2xy+y^2\) > cancel x^2+y^2 in both sides: \(2xy<2xy\) > reduce by 2: \(xy<xy\) > \(xy+xy>0\). Hope it's clear. \((xy)^2=(xy)^2=x^22xy+y^2\). \((x + y)^2=(x)^2+2x*y+(y)^2=x^2+2xy+y^2\).
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Re: Is x y < x + y?
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29 Jun 2013, 16:16
Is x – y < x + y?
(1) y < x
If y < x then x  y will always be positive. However, we don't know the signs for x and y so we don't know what x and y is. INSUFFICIENT
(2) xy < 0
This means that either x or y is negative and neither x or y are zero.
If we test for positive and negative values of x and y we will find that regardless of what x and y are (as long as one is positive and one is negative) the inequality will always fail (LHS and RHS will always equal one another) SUFFICIENT
(B)



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