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# Is |x – 3| > |y – 3|?

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Is |x – 3| > |y – 3|? [#permalink]

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15 Feb 2011, 13:28
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Question Stats:

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Is |x – 3| > |y – 3|?

(1) x > y.
(2) xу is not equal to 0.
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Jul 2013, 02:13, edited 2 times in total.

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Re: 180. Is Ix – 3I > Iy – 3I? (1) x > y. (2) xу is not equal [#permalink]

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15 Feb 2011, 13:59
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Is |x – 3| > |y – 3|?

x-3 > -(y-3)
x-3 > -y+3
x+y > 6
or
x-3 > y-3
x-y > 0
x>y

so; if x>y and x+y>6; we can be sure that |x-3| > |y-3|

(1) x > y.
But we don't know whether x+y>6. Not sufficient.

x=2
y=1
1<2

x=5
y=2
2>1

(2) xу is not equal to 0.
We don't know whether x>y or x+y>6. Not sufficient.
Same sample set from 1 can be used;

Together;
We don't know whether x+y>6. Not sufficient.

Same sample set from 1 can be used.

Ans: "E"
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Re: 180. Is Ix – 3I > Iy – 3I? (1) x > y. (2) xу is not equal [#permalink]

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15 Feb 2011, 14:05
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banksy wrote:
180. Is |x – 3| > |y – 3|?
(1) x > y.
(2) xу is not equal to 0.

Is |x – 3| > |y – 3|?

You need no algebra for this one. Question basically asks whether point x on the number line is further from 3 than point y (as |x-3| is the distance between points x and 3 on the number line and |y-3| is the distance between y and 3).

(1) x > y. Totally irrelevant.

(2) xу is not equal to 0. Also irrelevant, it just means that neither x nor y equals to zero.

(1)+(2) Two useless statements. Not sufficient.

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Re: Is |x – 3| > |y – 3|? [#permalink]

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23 Feb 2014, 05:05
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Bumping for review and further discussion.
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Re: Is |x – 3| > |y – 3|? [#permalink]

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07 May 2014, 21:14
Is |x – 3| > |y – 3|?

To make true the statement we have to alternatives:
x>y (both x and y positives)
x<y (both x and y negatives)

(1) x > y. If both are positive the answer will be YES but if they are negatives the answer will be NO
(2) xу is not equal to 0. This statement is irrelevant.

(1)+(2) We can not determine. Not sufficient.

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Re: Is |x – 3| > |y – 3|? [#permalink]

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04 Oct 2014, 21:06
E

Picking smart numbers is the best approach for this problem:
(1)
N: x=6, y=-6
Y: x=6, y=5
NS

(2)
Same numbers as (1)
NS

(1)+(2)
NS

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Re: Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 03:01
Hello Experts,
Since both sides of the eqn are positive I assumed we can approach the problem as below:
(x-3)^2 - (y-3)^2 >0 which leads to (x-3-y-3)(x-3-y+3)>0 and further (x-y-6)(x-y)>0 ie, we get 2 solutions : x>y or (x-y)>6.

Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.

Thanks & Regards,
Nab

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Re: Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 08:46
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Nab77 wrote:
Hello Experts,
Since both sides of the eqn are positive I assumed we can approach the problem as below:
(x-3)^2 - (y-3)^2 >0 which leads to (x-3-y-3)(x-3-y+3)>0 and further (x-y-6)(x-y)>0 ie, we get 2 solutions : x>y or (x-y)>6.

Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.

Thanks & Regards,
Nab

Hi Nab,
you have done two mistakes in the highlighted portion..

1) $$(x-3)^2 - (y-3)^2 >0.............. (x-3+y-3)(x-3(y-3))>0...... (x+y-6)(x-y)>0.....$$ and NOT (x-y-6)(x-y)>0

2) It is NOT x>y or (x-y)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6..
so two cases-
a) BOTH (x+y-6) and (x-y) are +ive or BOTh are -ive
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Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 09:42
chetan2u wrote:
Nab77 wrote:
Hello Experts,
Since both sides of the eqn are positive I assumed we can approach the problem as below:
(x-3)^2 - (y-3)^2 >0 which leads to (x-3-y-3)(x-3-y+3)>0 and further (x-y-6)(x-y)>0 ie, we get 2 solutions : x>y or (x-y)>6.

Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.

Thanks & Regards,
Nab

Hi Nab,
you have done two mistakes in the highlighted portion..

1) $$(x-3)^2 - (y-3)^2 >0.............. (x-3+y-3)(x-3(y-3))>0...... (x+y-6)(x-y)>0.....$$ and NOT (x-y-6)(x-y)>0

2) It is NOT x>y or (x-y)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6..
so two cases-
a) BOTH (x+y-6) and (x-y) are +ive or BOTh are -ive

Hi Chetan,
Oops yes I'm sorry that was a typo, i did get (x+y-6)(x-y)>0.
From 2) Do you mean that it is not an OR condition but an AND condition? I didn't understand that quite well, can you please explain again. Especially the part BUT x>y and (x+y)>6 or x<y and (x+y)<6. I didn't understand how the equality signs changed.

Regards,
Nab

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Re: Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 09:54
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Expert's post
Nab77 wrote:
chetan2u wrote:
Nab77 wrote:
Hello Experts,
Since both sides of the eqn are positive I assumed we can approach the problem as below:
(x-3)^2 - (y-3)^2 >0 which leads to (x-3-y-3)(x-3-y+3)>0 and further (x-y-6)(x-y)>0 ie, we get 2 solutions : x>y or (x-y)>6.

Because I saw Option 1 has one of these solutions ie.x>y I chose (A). Is it incorrect to choose an option if it is partially satisfied? Kindly help me understand.

Thanks & Regards,
Nab

Hi Nab,
you have done two mistakes in the highlighted portion..

1) $$(x-3)^2 - (y-3)^2 >0.............. (x-3+y-3)(x-3(y-3))>0...... (x+y-6)(x-y)>0.....$$ and NOT (x-y-6)(x-y)>0

2) It is NOT x>y or (x-y)>6 BUT x>y and (x+y)>6 or x<y and (x+y)<6..
so two cases-
a) BOTH (x+y-6) and (x-y) are +ive or BOTh are -ive

Hi Chetan,
Oops yes I'm sorry that was a typo, i did get (x+y-6)(x-y)>0.
From 2) Do you mean that it is not an OR condition but an AND condition? I didn't understand that quite well, can you please explain again. Especially the part BUT x>y and (x+y)>6 or x<y and (x+y)<6. I didn't understand how the equality signs changed.

Regards,
Nab

Hi,

the equation is-

$$(x+y-6)(x-y)>0$$...

The Left Hand Side can be >0 under two cases..

1) when both (x+y-6) and (x-y) are greater than 0... since Positive * Positive = Positive..
so x-y>0 or x>y................and x+y-6>0 or x+y>0.........
Example x = 5, y=2.. x>y and x+y>6 ... so (5+2-6)(5-2)>0.....1*3>0...YES

2) when both (x+y-6) and (x-y) are lesser than 0... since Negative * Negative = Positive..
so x-y<0 or x<y................and x+y-6<0 or x+y<6.........
x= 2 and y =3.....x<y and x+y<6 ... so (3+2-6)(2-3)>0.....(-1)(-1)>0..........1>0......YES
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 10:43
Quote:
Hi,

the equation is-

$$(x+y-6)(x-y)>0$$...

The Left Hand Side can be >0 under two cases..

1) when both (x+y-6) and (x-y) are greater than 0... since Positive * Positive = Positive..
so x-y>0 or x>y................and x+y-6>0 or x+y>0.........
Example x = 5, y=2.. x>y and x+y>6 ... so (5+2-6)(5-2)>0.....1*3>0...YES

2) when both (x+y-6) and (x-y) are lesser than 0... since Negative * Negative = Positive..
so x-y<0 or x<y................and x+y-6<0 or x+y<6.........
x= 2 and y =3.....x<y and x+y<6 ... so (3+2-6)(2-3)>0.....(-1)(-1)>0..........1>0......YES

Yes i realize where I went wrong!
Thanks a tonne for the explanation.

Thanks & Regards,
Nab

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Re: Is |x – 3| > |y – 3|? [#permalink]

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24 May 2016, 11:00
banksy wrote:
Is |x – 3| > |y – 3|?

(1) x > y.
(2) xу is not equal to 0.

(1) x > y.
Let's say x=4 and y=3, then |x – 3| > |y – 3|
but it x=-3 and y=-4, then the aboveineqzuality will not hold true.
Insufficient
(2) xу is not equal to 0
This statement means either x or y is not equal to 0. It can be +ve or -ve. not sufficient.

Combining both statements doesn't give a unique answer. Hence E is the answer
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Re: Is |x – 3| > |y – 3|? [#permalink]

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04 Oct 2017, 06:01
banksy wrote:
Is |x – 3| > |y – 3|?

(1) x > y.
(2) xу is not equal to 0.

DS: Is |x – 3| > |y – 3|?

Statement 1 : x>y
Lets say x =6 , y = 4 |x-3| = 3>|y-3| = 1
Lets say x = 1, y = -4 |x-3|= 2<|y-3| = 7

NOT SUFFICIENT

Statement 2 : xy is not equal to 0 . So, neither x nor y is not equal to 0 .
Lets say x =6 , y = 4 |x-3| = 3>|y-3| = 1
Lets say x = 1, y = -4 |x-3|= 2<|y-3| = 7
NOT SUFFICIENT

Combined : same examples
NOT SUFFICIENT

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Re: Is |x – 3| > |y – 3|? [#permalink]

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05 Oct 2017, 21:54
Is |x – 3| > |y – 3|?

(1) x > y.
(2) xу is not equal to 0

2. XY not equal to 0 we don't know the value of X and Y so Insufficient

2. If x =10 and Y = -100 then |Y-3| is greater if x=10 and Y= 5 then |X-3| is greater so insuffiecient

together the same example holds good, not sufficient hence Answer should be E

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Re: Is |x – 3| > |y – 3|?   [#permalink] 05 Oct 2017, 21:54
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