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x and y..........

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milind1979
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x and y.......... [#permalink] New post   03 May 2009, 21:06
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A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Re: x and y.......... [#permalink] New post   03 May 2009, 21:59
B is correct.

Y >= 0 means y is not negative.

1. lx-3l >= y
In this case, x could have any (-ve, 0 or +ve) value. Its not sufficient.

2. lx-3l =< -y

In this case, lx-3l is always equal to 0 or +ve. Since y is >=0, y cannot be +ve to have statement 2 to be true.
So lx-3l must be 0 and hence x = 3. Its sufficient.

B.
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Re: x and y.......... [#permalink] New post   04 May 2009, 04:53
Yes, It is answer B.


(1) |x-3| ≥ y

This statement will hold true for several values of x and y.
for y = 0, x = to
for y = 1, x = ( to 2) and (4 to )
for y = 2, x = ( to 1) and (5 to )
...
and so on


(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value.
example,
for y = 1, |x-3| can never be less than -y=-1.
for y = 2, |x-3| can never be less than -y=-2.
-y ≠ +ve value, because of precondition that y ≥ 0
The only value of y, for which this statement is true is y = 0.
|x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.


hope it helps.
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Re: x and y.......... [#permalink] New post   04 May 2009, 08:13
Agree on B.
x can only be 0
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Re: x and y.......... [#permalink] New post   04 May 2009, 22:22
goldeneagle94 wrote:
Yes, It is answer B.


(1) |x-3| ≥ y

This statement will hold true for several values of x and y.
for y = 0, x = to
for y = 1, x = ( to 2) and (4 to )
for y = 2, x = ( to 1) and (5 to )
...
and so on


(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value.
example,
for y = 1, |x-3| can never be less than -y=-1.
for y = 2, |x-3| can never be less than -y=-2.
-y ≠ +ve value, because of precondition that y ≥ 0
The only value of y, for which this statement is true is y = 0.
|x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.


hope it helps.


Understood why B.

Can you explain how to solve general equation type l something l > something else? please explain this for statement 1
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Re: x and y.......... [#permalink] New post   05 May 2009, 03:41
bandit wrote:
goldeneagle94 wrote:
Yes, It is answer B.


(1) |x-3| ≥ y

This statement will hold true for several values of x and y.
for y = 0, x = to
for y = 1, x = ( to 2) and (4 to )
for y = 2, x = ( to 1) and (5 to )
...
and so on


(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value.
example,
for y = 1, |x-3| can never be less than -y=-1.
for y = 2, |x-3| can never be less than -y=-2.
-y ≠ +ve value, because of precondition that y ≥ 0
The only value of y, for which this statement is true is y = 0.
|x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.


hope it helps.


Understood why B.

Can you explain how to solve general equation type l something l > something else? please explain this for statement 1



Most of the time, values have to be substituted in the variable to determine its RANGE. Like in Statement 1, we found that the range is ANY value.
for example, |x+4| > 0, then range of x will be -4 < x <

Try to figure out its RANGE.
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Re: x and y.......... [#permalink] New post   05 May 2009, 18:25
Hi

Can you explain the below step alone.

|x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true

Are we coming to tell x = 3 or x = 0.

i understood |x-3| ≤ 0 . A mod can never be less than or equal to a negative value. so | X-3| =0 and x =3.

Please correct me if i am wrong.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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gmatclubot
Re: x and y.......... [#permalink]
05 May 2009, 18:25
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