absolute

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 484
Joined: Sun Jul 30, 2006 7:01 pm
Thanked: 2 times
Followed by:1 members

absolute

by magical cook » Mon Dec 10, 2007 10:04 am
if y>=0, what is the value of x?

1) lx-3l >= y

2) lx-3l <= -y

Master | Next Rank: 500 Posts
Posts: 214
Joined: Wed Nov 14, 2007 6:30 am
Thanked: 15 times

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

Master | Next Rank: 500 Posts
Posts: 158
Joined: Mon Dec 03, 2007 8:32 am
Thanked: 7 times

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.

Master | Next Rank: 500 Posts
Posts: 158
Joined: Mon Dec 03, 2007 8:32 am
Thanked: 7 times

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.

Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Aug 23, 2007 5:11 am
Thanked: 2 times

by gmatguy16 » Mon Dec 10, 2007 3:50 pm
can someone confirm oa?

Master | Next Rank: 500 Posts
Posts: 484
Joined: Sun Jul 30, 2006 7:01 pm
Thanked: 2 times
Followed by:1 members

by magical cook » Mon Dec 10, 2007 4:26 pm
Sorry for the late reply - OA is B. :)

Senior | Next Rank: 100 Posts
Posts: 62
Joined: Tue Jan 16, 2007 9:40 am
Location: Arlington,VA

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 ?

Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

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

Master | Next Rank: 500 Posts
Posts: 214
Joined: Wed Nov 14, 2007 6:30 am
Thanked: 15 times

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

User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pm
Location: Silicon valley, California
Thanked: 30 times
Followed by:1 members

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

Master | Next Rank: 500 Posts
Posts: 214
Joined: Wed Nov 14, 2007 6:30 am
Thanked: 15 times

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