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absolute

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by sujaysolanki » Mon Dec 10, 2007 10:44 am
From 1 we have 2 possibilities

|x| = x or -x so two values

From 2 also the same thing

Combining we have -x + 3 >=y
i.e. x + y >=3 ... so i think E
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by StarDust845 » Mon Dec 10, 2007 1:58 pm
Answer is B.

|x-3| <y>= 0

if y = 0, then x MUST be 3.

if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.


Calista.
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by StarDust845 » Mon Dec 10, 2007 2:00 pm
Correcting some typos in my earlier post.

Answer is B.

From 2) |x-3| <y>= 0

if y = 0, then x MUST be 3.

if y is say 4 then |x-3| <= -4 there is no value for x. So 2) can't be true. But it is given that |x-3| <= -y.. hence y must be 0 . so x must be 3.


Calista.
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by gmatguy16 » Mon Dec 10, 2007 3:50 pm
can someone confirm oa?
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by magical cook » Mon Dec 10, 2007 4:26 pm
Sorry for the late reply - OA is B. :)
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by GMATDUD » Mon Dec 10, 2007 8:16 pm
I still did not get the explanation .

B) lx-3l <= -y

|x|<= -Y+3

All this says is x is negative and 3 more than y. Absolute val is my weak area. Can someone help by explaining in more detail ?
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by samirpandeyit62 » Tue Dec 11, 2007 2:35 am
Hi GMATDUD, I would solve for stmt 2 like this

if y>=0, what is the value of x?

B) lx-3l <y>=0 so -y will be 0 or -ve

now |x-3| is an absolute value so it will be always +ve

so in the given inequality the only real world case that will arise will be
when y=0 as for other values of y -y will be -ve which cannot be more than a +ve value (|x-3|)

so we have |x-3| = 0

or x =3

B
Regards
Samir
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by sujaysolanki » Tue Dec 11, 2007 4:48 am
Could someone explain ..i still dont get it
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Re: absolute

by jayhawk2001 » Tue Dec 11, 2007 11:51 am
magical cook wrote:if y>=0, what is the value of x?

1) lx-3l >= y

2) lx-3l <y>= 0.
Since y >=0, only value of y that satisfies this is y=0.

So |x-3| <= 0. So, we can find x.

Hence B
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by sujaysolanki » Tue Dec 11, 2007 9:10 pm
makes sense thanks ..
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