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DS mod

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Director
Joined: 20 Sep 2006
Posts: 653

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29 Sep 2008, 19:07
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SVP
Joined: 17 Jun 2008
Posts: 1529

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29 Sep 2008, 21:02
C.

What the two statements are really saying is that mod of (x-3) will be absolute positive and that positive number is greater than y and less than -y. This is possible only when y = 0 .

Now, if we re-write the two statements then
|x-3| >= 0 and |x-3| <=0

Again, both the statements will be true only if |x-3| = 0 and that means x = 3.
Intern
Joined: 29 Sep 2008
Posts: 46

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30 Sep 2008, 15:38
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3
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Joined: 18 May 2008
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30 Sep 2008, 16:33
I think we dnt need 2 combine the two equations.
S1 says that |x-3| >= 0 which is can give either x=3 or x>3 Insufficient
S2 says that |x-3| <=0 which is only possible when x=0 because modulus cn nev be less than 0
So I think B should be the ans

scthakur wrote:
C.

What the two statements are really saying is that mod of (x-3) will be absolute positive and that positive number is greater than y and less than -y. This is possible only when y = 0 .

Now, if we re-write the two statements then
|x-3| >= 0 and |x-3| <=0

Again, both the statements will be true only if |x-3| = 0 and that means x = 3.

Last edited by ritula on 01 Oct 2008, 15:44, edited 1 time in total.
SVP
Joined: 17 Jun 2008
Posts: 1529

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01 Oct 2008, 11:01
I got what you are saying. Question already gives clue that y >=0 and hence B should be sufficient.

aim2010 wrote:
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3
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Joined: 07 Nov 2007
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01 Oct 2008, 11:48
rao_1857 wrote:
Attachment:
m_2.JPG

|x-3|<= -y (Y is positve or zero)

|x-3| --> must be postive or zero.
-y is always -ve

so only possibility is |x-3|=0 -->x=3

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Director
Joined: 20 Sep 2006
Posts: 653

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02 Oct 2008, 05:46
aim2010 wrote:
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3

Thanks aim2010 thanks for clarifying ... OA is B
Re: DS mod   [#permalink] 02 Oct 2008, 05:46
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