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# If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y

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If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]  21 Jan 2012, 08:57
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If y >= 0, What is the value of x?

(1) |x-3| >= y
(2) |x-3| <= -y

[Reveal] Spoiler:
I solved this way -

1) |x-3| >= y

x -3 >=y
x >= y + 3

or

3 - x >=y
x <= 3 - y

so y + 3 <= x <= 3 - y

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y
x <=3-y

or

3-x <=-y
x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

But the OA is different.
[Reveal] Spoiler: OA
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Re: DS inequalities [#permalink]  21 Jan 2012, 09:23
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Apex231 wrote:
if y >= 0, What is the value of x?
1) |x-3| >= y
2) |x-3| <= -y

I solved this way -

1) |x-3| >= y

x -3 >=y
x >= y + 3

or

3 - x >=y
x <= 3 - y

so y + 3 <= x <= 3 - y

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y
x <=3-y

or

3-x <=-y
x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

But the OA is different.

If y\geq{0}, what is the value of [b]x?[/b]

(1) |x - 3|\geq{y}. As given that y is non negative value then |x - 3| is more than (or equal to) some non negative value, (we could say the same ourselves as absolute value in our case (|x - 3|) is never negative). So we can not determine single numerical value of x. Not sufficient.

Or another way: to check |x - 3|\geq{y}\geq{0} is sufficient or not just plug numbers:
A. x=5, y=1>0, and B. x=8, y=2>0: you'll see that both fits in |x - 3|>=y, y\geq{0}.

Or another way:
|x - 3|\geq{y} means that:

x - 3\geq{y}\geq{0} when x-3>0 --> x>3

OR (not and)
-x+3\geq{y}\geq{0} when x-3<0 --> x<3

Generally speaking |x - 3|\geq{y}\geq{0} means that |x - 3|, an absolute value, is not negative. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|\leq{-y}. Now, as |x-3| is never negative (0\leq{|x-3|}) then 0\leq{-y} --> y\leq{0} BUT stem says that y\geq{0} thus y=0. |x-3|\leq{0} --> |x-3|=0=y (as absolute value, in our case |x-3|, can not be less than zero) --> x-3=0 --> x=3. SUFFICIENT

In other words:
-y is zero or less, and the absolute value (|x-3|) must be at zero or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

There is a following problem with your solution:
If x<3 --> -(x-3)\geq{y} --> 3-y\geq{x};
OR:
If x\geq{3} --> (x-3)\geq{y} --> x\geq{3+y};

But you can not combine these inequalities and write: 3+y\leq{x}\leq{3-y} as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

Also discussed here: if-y-geq-0-what-is-the-value-of-x-1-x-3-geq-y-91640.html

Hope it helps..
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Re: DS inequalities [#permalink]  21 Jan 2012, 09:50
Bunuel wrote:
But you can not combine these inequalities and write: 3+y\leq{x}\leq{3-y} as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

Thanks Bunuel..realize my mistake now...great explanation as always...
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Gmat prep inequalities [#permalink]  31 Jan 2012, 04:40
Hi friends can anyone please explain this.....
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Re: Gmat prep inequalities [#permalink]  31 Jan 2012, 04:44
kotela wrote:
Hi friends can anyone please explain this.....

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Re: Gmat prep inequalities   [#permalink] 31 Jan 2012, 04:44
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