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(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.

okay so my answer is wrong..can't figure out what i did wrong on statement 2

What you missed is that (2) is true not only when x>y but also when x=y. Hence we can not say for sure that x>y, as x>=y. Not sufficient.
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\(|y - x| = x - y\). 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 \(x>=y\)

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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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 >

Last edited by chix475ntu on 11 Feb 2010, 10:06, edited 1 time in total.

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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1) Now x=(3/4)y so y > x means statement 1 alone is sufficient to answer the question.

2) Using some values for x & y we can find if x>y then statement |y - x| = x - y is true and if y>x then statement |y - x| = x - y is not true. Hence statement 2 alone is sufficient to answer question.

Either statement is sufficient to answer question so answer is D.

(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.

(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.

(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 \(y-x\leq{0}\).
_________________

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]

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27 Nov 2013, 06:13

Economist wrote:

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) |y - x| = x - y

Question is: Is x>y?

Statement 1 4x = 3y x/y=3/4 Not sufficient, we need to know their signs

Statement 2 |y - x| = x - y We can rearrange LHS as |-(-y + x|) = |(x-y|) So we end up with |(x-y|=x-y, meaning that |x-y|>=0 or x>=y Could be equal or could be greater. Insuff

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

1

This post received KUDOS

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

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]

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09 Aug 2016, 23:33

1

This post was BOOKMARKED

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.

(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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