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If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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 12 Nov 2009, 09:45
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If xy ≠ 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
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If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y This is a good one. +1 Economist for it. If xy ≠ 0, is x > y?Question asks whether \(x>y\)? (1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x<y\), BUT when they are both negative \(y<x\). Not sufficient. (2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient. (1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient. Answer: C.
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This is a good question, simple with a small trap
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If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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12 Nov 2009, 10:26
|y - x| = x - y
y - x = x - y
2x = 2y
x=y so answer is no
-y + x = x-y
0=0
answer is no
sufficient
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lagomez wrote:
2. |y - x| = x - y
y - x = x - y
2x = 2y
x=y so answer is no
-y + x = x-y
0=0
answer is no
sufficient
One more thing: the red part is not correct. \(|y - x| = x - y\). Look at the LHS it's absolute value, it's never negative, hence RHS or \(x-y\) is never negative, \(x-y>=0\) or \(y-x<=0\). If so then only one possibility for \(|y - x|\), it must be \(-y+x\). So we'll have: Condition: \(y-x<=0\), which is the same as \(x>=y\) And: \(|y - x| = x - y\)--> \(-y+x=x-y\), --> \(0=0\). The above means that \(|y - x| = x - y\) is always true for \(x>=y\). But as we concluded earlier it's not enough. We need to be sure that \(x>y\). And \(x>=y\) leaves the possibility that they are equal, which is removed with the statement (1) when considering together. Hence C.
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Re: Inequalities [#permalink]
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Updated on: 11 Feb 2010, 10:06
st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value all the cases as cay that x>y but fails when x=y combining we can answer the question C <didnt consider x=y option, so answered as B  >
Originally posted by chix475ntu on 10 Feb 2010, 22:16.
Last edited by chix475ntu on 11 Feb 2010, 10:06, edited 1 time in total.
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Re: Inequalities [#permalink]
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11 Feb 2010, 00:08
chix475ntu wrote: st 1) y=1, x=3/4 - false
y=-1, x=-3/4 - true
Not sufficient
st 2)
|y-x| =x-y--> x>y
Sufficient
coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value
B B alone is not sufficient as we dont know whether x=y? So A gives information that x is not = y. So Final ans is C This was nice question...even I would have done the same mistake in the exam..Its really important to think of all the possibilities on the G day.
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Re: Inequalities [#permalink]
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11 Feb 2010, 11:55
Given: \(xy\neq0\). Question: is \(x>y\) true? (1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\). (2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient. (1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient. Answer: C. Hope it helps.
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Re: Inequalities [#permalink]
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30 Sep 2013, 14:42
Bunuel wrote: Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps. Dear Bunuel: I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one.. I saw a user on this thread was solving the second statement as this: 2. |y - x| = x - y y - x = x - y 2x = 2y x=y -y + x = x-y 0=0 Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem. If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me. Thanks in advance.
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Re: Inequalities [#permalink]
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01 Oct 2013, 00:34
Aidhen wrote: Bunuel wrote: Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps. Dear Bunuel: I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one.. I saw a user on this thread was solving the second statement as this: 2. |y - x| = x - y y - x = x - y 2x = 2y x=y -y + x = x-y 0=0 Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem. If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me. Thanks in advance. There can be many correct ways to solve a questions. Try this one: (2) says that \(|y-x|=-(y-x)\). We know that \(|x|=-x\), when \(x\leq{0}\), thus \(y-x\leq{0}\).
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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20 Oct 2014, 22:26
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If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
Consider (1)
4x=3y ; x=3/-3 and y =4/-4 satisfy this condition. so x can be > y or x can be <y.
insufficient
Consider (2)
|y - x| = x - y
x= 4 y =3 satisfy this , and x>y
x= 3 y=3 satisfy this but x =y
x= 3 and y = 4 doesnt satisfy
x=-3 and y =-4 satisfy this and x >y
hence insufficient
combining 1 and 2 the only option that satisfy 1 and 2 is x=-3 and y=-4 and hence x>y
Therefore C is the answer.
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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Bunuel wrote: Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps. Hello Bunuel, Thank you for your answer. Although , as per mod property, isn't it the case that |x| = x when x>=0 and -x when x<0 ? I would like to know why you are considering |x| = -x when x<=0, because this is where the answer is really hinged on. Thanks!
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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10 Aug 2016, 00:58
ramblersm wrote: Bunuel wrote: Given: \(xy\neq0\). Question: is \(x>y\) true?
(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).
(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.
(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.
Answer: C.
Hope it helps. Hello Bunuel, Thank you for your answer. Although , as per mod property, isn't it the case that |x| = x when x>=0 and -x when x<0 ? I would like to know why you are considering |x| = -x when x<=0, because this is where the answer is really hinged on. Thanks! |0| = 0 = -0, so you can include = sign in either of the cases: |x| = -x, when x<=0 and |x| = x, when x >= 0.
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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31 Jan 2018, 08:28
Bunuel wrote: Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y This is a good one. +1 Economist for it. If xy ≠ 0, is x > y?Question asks whether \(x>y\)? (1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x>y\), BUT when they are both negative \(y>x\). Not sufficient. (2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient. (1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient. Answer: C. For (1), isn't it the opposite? when both +ve x<y, when both -ve x>y.
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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31 Jan 2018, 08:36
urvashis09 wrote: Bunuel wrote: Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y This is a good one. +1 Economist for it. If xy ≠ 0, is x > y?Question asks whether \(x>y\)? (1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x>y\), BUT when they are both negative \(y>x\). Not sufficient. (2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient. (1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient. Answer: C. For (1), isn't it the opposite? when both +ve x<y, when both -ve x>y. Yes but that doesn't change the answer anyway.
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If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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05 Feb 2018, 06:59
Kudos for the explanation. Am having trouble with absolute value and inequalities. How can I improve????
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
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