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# is z ]<1? =absolute value 1) z>-1 2) z<1

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Director
Joined: 17 Oct 2005
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is z ]<1? =absolute value 1) z>-1 2) z<1 [#permalink]  15 Jan 2006, 02:18
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is [z^2/z]<1? [] =absolute value

1) z>-1
2) z<1

Last edited by joemama142000 on 15 Jan 2006, 04:01, edited 1 time in total.
Director
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

Re: absolute value DS [#permalink]  15 Jan 2006, 02:42
joemama142000 wrote:
is [z^2/x]<1? [] =absolute value

1) z>-1
2) z<1

My mistake, should be.

is [z^2/z]<1? [] =absolute value

1) z>-1
2) z<1
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
Followers: 1

Kudos [?]: 26 [0], given: 0

A) is not enough to solve this. Z could be bigger or less than 1
B) is also insufficient , for same reason
Both together fix value for z which is less than 1, as absolute value. IMO it is C)
VP
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 30 [0], given: 0

Re: absolute value DS [#permalink]  15 Jan 2006, 07:20
joemama142000 wrote:
is lz^2/zl<1?
1) z>-1
2) z<1

C.
from i, z could be less than 1 or more than 1. so not enough.
from ii, z could be less than -1 or more than -1. so not enough.
togather z is in between -1 and 1. so suff.
Manager
Joined: 15 Aug 2005
Posts: 136
Followers: 2

Kudos [?]: 10 [0], given: 0

Re: absolute value DS [#permalink]  15 Jan 2006, 13:16
joemama142000 wrote:
is [z^2/z]<1? [] =absolute value

1) z>-1
2) z<1

C
|z^2/z| < 1
=> (z + sqrt(2) ) (z - sqrt(2) ) < 0

so neither (i) nor (ii) by themselves are sufficient. But together they are. C
Director
Joined: 17 Oct 2005
Posts: 940
Followers: 1

Kudos [?]: 62 [0], given: 0

oa is C

believe2, how did y ou manipulate this equation?
Manager
Joined: 15 Aug 2005
Posts: 136
Followers: 2

Kudos [?]: 10 [0], given: 0

Re: absolute value DS [#permalink]  15 Jan 2006, 15:54
believe2 wrote:
joemama142000 wrote:
is [z^2/z]<1? [] =absolute value

1) z>-1
2) z<1

C
|z^2/z| < 1
=> (z + sqrt(2) ) (z - sqrt(2) ) < 0

so neither (i) nor (ii) by themselves are sufficient. But together they are. C

Oh! my mistake....the z & 2 looked similar
I solved |z^2/2| < 1 rather than |z^2/z| < 1 ...... lucky to get the right ans. with the wrong method!!
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 183 [0], given: 0

1) z>-1, so z an be integer values like 0,1,2 or fractions like 1/2 etc.

If z = 0 -> undefined
If z = 1, |z^2/z|=1
If z =2, |z^2/z| > 1
If z = 1/2, |z^2/z| < 1

Insufficient.

2) z<1, so z can be 0,-1,-2, or fractions like -1/2

If z = 0 -> undefined
If z = -1, |z^2/z|=1
If z = -2, |z^2/z|>1
If z = -1/2, |z^2/z| < 1.

Insufficient.

If -1<z<1, then z can only be fractions such as 1/2, 1/4, -1/8 etc. In these cases, |z^2/z| is always less than 1.

Ans C
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