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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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12 Nov 2009, 09:45
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If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) |y - x| = x - y
[Reveal] Spoiler: OA

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Re: xy [#permalink]

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12 Nov 2009, 10:05
Economist wrote:
If xy ≠ 0, is x > y?

(1) 4x = 3y

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

I'm getting B

1. 4x = 3y
4/3x = y

if x = 3 y = 4 so yes
if x = -3 y = - 4 so no

not sufficient

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

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Re: xy [#permalink]

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12 Nov 2009, 10:08
Economist wrote:
If xy ≠ 0, is x > y?

(1) 4x = 3y

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

B

1. 4x=3y
if we take x=1 then y = 4/3 and we get x>y as No
if we take x=-3 then y= -4 then x>y is Yes
hence insuff

2. Considering x and y have different value,mod value will always be +ve so x-y = some +ve number.hence x>y. Suff

Last edited by kp1811 on 12 Nov 2009, 10:12, edited 1 time in total.

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Re: xy [#permalink]

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12 Nov 2009, 10:11
kp1811 wrote:
Economist wrote:
If xy ≠ 0, is x > y?

(1) 4x = 3y

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

B

1. 4x=3y
if we take x=1 then y = 4/3 and we get x>y as No
if we take x=-3 then y= -4 then x>y is Yes
hence insuff

2. mod value will always be +ve so x-y = some +ve number.hence x>y. Suff

i got the answer but where is my logic off on statement 2?
what if x = 3 and y = 3?

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Re: xy [#permalink]

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12 Nov 2009, 10:14
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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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Re: xy [#permalink]

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12 Nov 2009, 10:16
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This is a good question, simple with a small trap

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Re: xy [#permalink]

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12 Nov 2009, 10:16
okay so my answer is wrong..can't figure out what i did wrong on statement 2

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Re: xy [#permalink]

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12 Nov 2009, 10:22
lagomez wrote:

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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Re: xy [#permalink]

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12 Nov 2009, 10:26
Bunuel wrote:
lagomez wrote:

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. Nor sufficient.

so what is the best way to approach a statement like this?

i have an absolute value and an equal sign but the question is asking for greater than or less than

that always throws me off because had it been |y-x| > x-y would have been easier but that equal sign throws me off

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Re: xy [#permalink]

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12 Nov 2009, 10:32
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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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10 Feb 2010, 22:16
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.

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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, 09:00
dmetla wrote:
If xy ≠ 0, is x > y?

(1) 4x = 3y

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

Neither of x or y is zero as xy≠0.

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.

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Re: Inequalities [#permalink]

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11 Feb 2010, 11:55
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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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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

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

Answer is C
Hope it helps
Cheers!
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]

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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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09 Aug 2016, 23:33
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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] 10 Aug 2016, 00:58

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