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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.

|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.

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, 09:06, 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.

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.

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
26 Sep 2013, 08:28

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(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]
27 Nov 2013, 05:13

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

Both together, they can't be equal to it is Suff

Answer is C Hope it helps Cheers! J

gmatclubot

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
27 Nov 2013, 05:13