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

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Manager
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x and y.......... [#permalink]

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New post 03 May 2009, 21:06
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A
B
C
D
E

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Kudos [?]: 104 [0], given: 1

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Re: x and y.......... [#permalink]

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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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Manager
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Re: x and y.......... [#permalink]

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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 = Image to Image
for y = 1, x = (Image to 2) and (4 to Image)
for y = 2, x = (Image to 1) and (5 to Image)
...
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
Image 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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Manager
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Re: x and y.......... [#permalink]

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New post 04 May 2009, 08:13
Agree on B.
x can only be 0

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Re: x and y.......... [#permalink]

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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 = Image to Image
for y = 1, x = (Image to 2) and (4 to Image)
for y = 2, x = (Image to 1) and (5 to Image)
...
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
Image 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

Kudos [?]: 149 [0], given: 10

Manager
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Re: x and y.......... [#permalink]

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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 = Image to Image
for y = 1, x = (Image to 2) and (4 to Image)
for y = 2, x = (Image to 1) and (5 to Image)
...
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
Image 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 < Image

Try to figure out its RANGE.

Kudos [?]: 60 [0], given: 3

Senior Manager
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Re: x and y.......... [#permalink]

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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.

Kudos [?]: 176 [0], given: 5

Re: x and y..........   [#permalink] 05 May 2009, 18:25
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