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by eaakbari » Thu Apr 22, 2010 9:58 am
IMO B

Do confirm
Last edited by eaakbari on Thu Apr 22, 2010 10:21 am, edited 1 time in total.
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by eaakbari » Thu Apr 22, 2010 10:02 am
(1)
|x-3|>=y
buy y>= 0
hence |x-3|>=0
Insuff

(2)
|x-3| <= -y
but y>=0
therefore -y<=0
Hence |x-3| <= 0
Absolute value cannot be negative Hence x=3

Suff

Answer B
Last edited by eaakbari on Thu Apr 22, 2010 10:21 am, edited 1 time in total.
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by tpr-becky » Thu Apr 22, 2010 10:14 am
statement one simply says the absolute value of x-3 is positive or zero- but we know that already because an absolute value is always positive so that will give no information regarding the value of x.

Statement 2 says that the absolute value of x-3 is less than or equal to either a negative or a zero - an absolute value can never be less than a negative number so tha tmeans that x-3 must equal zero. that means x=3

So IMO the answer is B.
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by eaakbari » Thu Apr 22, 2010 10:17 am
tpr-becky wrote:statement one simply says the absolute value of x-3 is positive or zero- but we know that already because an absolute value is always positive so that will give no information regarding the value of x.

Statement 2 says that the absolute value of x-3 is less than or equal to either a negative or a zero - an absolute value can never be less than a negative number so tha tmeans that x-3 must equal zero. that means x=3

So IMO the answer is B.
Ooh Thanks Becky, I did make a silly mistake. Shall edit my post. Thanks
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by iamseer » Thu Apr 22, 2010 10:41 am
Thanks eaakbari and tpr-becky. That was quick and very well explained.
I was thinking on a totally different line :)
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by iamseer » Thu Apr 22, 2010 10:56 am
tpr-becky wrote:statement one simply says the absolute value of x-3 is positive or zero- but we know that already because an absolute value is always positive so that will give no information regarding the value of x.

Statement 2 says that the absolute value of x-3 is less than or equal to either a negative or a zero - an absolute value can never be less than a negative number so tha tmeans that x-3 must equal zero. that means x=3

So IMO the answer is B.
Please if could help me with my line of thinking too....

I was thinking
for 1.
|x-3| >= y therefore x-3>=y (for x>=3) or 3-x>=y (for x<3). i.e x>=y+3 or x<=3-y. Now b'cos y can take any value >=0, x can take more than one values. INSUFFICIENT

for 2.
|x-3| <= -y therefore x-3>=-y (for x>=3) or 3-x>= -y (for x<3). i.e x>=3-y or x<=3+y. Now again b'cos y can take any value >=0, x can take more than one values.

Please correct me where I am going wrong.
Thanks.
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by eaakbari » Thu Apr 22, 2010 11:14 am
iamseer wrote:
tpr-becky wrote:statement one simply says the absolute value of x-3 is positive or zero- but we know that already because an absolute value is always positive so that will give no information regarding the value of x.

Statement 2 says that the absolute value of x-3 is less than or equal to either a negative or a zero - an absolute value can never be less than a negative number so tha tmeans that x-3 must equal zero. that means x=3

So IMO the answer is B.
Please if could help me with my line of thinking too....

I was thinking
for 1.
|x-3| >= y therefore x-3>=y (for x>=3) or 3-x>=y (for x<3). i.e x>=y+3 or x<=3-y. Now b'cos y can take any value >=0, x can take more than one values. INSUFFICIENT

for 2.
|x-3| <= -y therefore x-3>=-y (for x>=3) or 3-x>= -y (for x<3). i.e x>=3-y or x<=3+y. Now again b'cos y can take any value >=0, x can take more than one values.

Please correct me where I am going wrong.
Thanks.
The deal here is you cannot open a modulus the way you did. You can square it and open it, but that wont help in this question.
For any |x| = 1
Implies x=1 or x=-1

Also for a modulus operator
for |x|
= x if x>0
=-x if x<0


HTH
Last edited by eaakbari on Fri Apr 23, 2010 11:18 am, edited 1 time in total.
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by tpr-becky » Thu Apr 22, 2010 2:50 pm
You cannot have an absolute value equal to a negative number, it is always positive - the way you have opened the equation is not allowed because it does not eliminate the possiblity of the absolute value being negative.
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by iamseer » Fri Apr 23, 2010 11:10 am
eaakbari wrote: For any |x| = 1
Implies -1<x<1
sorry didn't get that. B'cos that would mean |0.6| = 1 ????
I think, |x| means distance from zero. So, if |x| = 1 then x=1 or x=-1, right?
eaakbari wrote: Also for a modulus operator
for |x|
= x if x>0
=-x if x<0
HTH
Yes, this is correct.
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by iamseer » Fri Apr 23, 2010 11:16 am
tpr-becky wrote:You cannot have an absolute value equal to a negative number, it is always positive - the way you have opened the equation is not allowed because it does not eliminate the possiblity of the absolute value being negative.
Yups now I get it. Thanks for the explanation.
So, b'cos they have specifically mentioned y>=0, -y is always negative or zero. And absolute value cannot be negative so for the absolute value |x-3| to be equal or less than y only possibility is when y=0.

my wrong: in my working I was not taking the fact y>=0 into account and in the process making things unnecessarily difficult :)
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by eaakbari » Fri Apr 23, 2010 11:18 am
iamseer wrote:
eaakbari wrote: For any |x| = 1
Implies -1<x<1
sorry didn't get that. B'cos that would mean |0.6| = 1 ????
I think, |x| means distance from zero. So, if |x| = 1 then x=1 or x=-1, right?
eaakbari wrote: Also for a modulus operator
for |x|
= x if x>0
=-x if x<0
HTH
Yes, this is correct.
My bad, Dont know how i mistyped that. I meant x=-1 or x=1. Thanks for correcting. Shall edit the post
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