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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] New post 12 Nov 2009, 08:45
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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] New post 12 Nov 2009, 09:05
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] New post 12 Nov 2009, 09:08
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, 09:12, edited 1 time in total.
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Re: xy [#permalink] New post 12 Nov 2009, 09:11
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] New post 12 Nov 2009, 09:14
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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] New post 12 Nov 2009, 09:16
This is a good question, simple with a small trap
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Re: xy [#permalink] New post 12 Nov 2009, 09:16
okay so my answer is wrong..can't figure out what i did wrong on statement 2
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Re: xy [#permalink] New post 12 Nov 2009, 09:22
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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] New post 12 Nov 2009, 09:26
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] New post 12 Nov 2009, 09:32
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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] New post 10 Feb 2010, 21:16
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 :ouch >

Last edited by chix475ntu on 11 Feb 2010, 09:06, edited 1 time in total.
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Re: Inequalities [#permalink] New post 10 Feb 2010, 23:08
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] New post 11 Feb 2010, 08:00
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] New post 11 Feb 2010, 10:55
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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: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink] New post 26 Sep 2013, 08:28
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Re: Inequalities [#permalink] New post 30 Sep 2013, 13:42
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] New post 30 Sep 2013, 23:34
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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] New post 03 Oct 2013, 01:59
This I why i am thinking of:

1) Gives us X=3/4Y. But it does not help us to see if X>Y (try some numbers to try)
2) give us X>Y or X = Y !!

Therefore we need both statements (Answer C) to answer the question!

Is it the OA?
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink] New post 03 Oct 2013, 02:02
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Paris75 wrote:
This I why i am thinking of:

1) Gives us X=3/4Y. But it does not help us to see if X>Y (try some numbers to try)
2) give us X>Y or X = Y !!

Therefore we need both statements (Answer C) to answer the question!

Is it the OA?


Yes, the OA is C. It's given under the spoiler in the original post.
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Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink] New post 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 :)
Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y   [#permalink] 27 Nov 2013, 05:13
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