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Director
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Is n < 1 ? (1) (n^x) n < 0 (2) x^ 1 = 2 OA later.. [#permalink]
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08 Jul 2007, 10:46
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Is n < 1 ?
(1) (n^x) – n < 0
(2) x^–1 = –2
OA later...



Director
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Re: DS Modulus [#permalink]
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08 Jul 2007, 12:37
vshaunak@gmail.com wrote: Is n < 1? (1) (n^x) – n < 0 (2) x–1 = –2 OA later...
Should be E.
from 1:
n^x – n < 0
n^x < n
if n = 4, and x = 1/2, (n^x) < (n) and n is grater than 1
if n = 2, and x = 1, (n^x) < (n) and n is grater than 1
if n = 0.25, and x = 2, (n^x) < (n) and n is smaller than 1. so nsf.
from 2:
x – 1 = –2
x = – 1. also nsf.
From 1 and 2:
if x = 1 and n = 2, (n^x) < (n) and n is grater than 1
if x = 1 and n = 0.5, (n^x) < (n) and n is smaller than 1
nsf.



Director
Joined: 14 Jan 2007
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Think more. OA is not 'E'



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Joined: 07 Jul 2004
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Location: Singapore

Try st2 first:
x1 = 2
x = 1 > nothing about n, insufficient.
try st1:
If x is even, say 2, then:
n^2n < 0 if n is a positive integer
n^2n < 0 if n is a positive fraction
If x is odd, say 3, then:
n^3n < 0 if n is a negative integer
n^3n < 0 if n is a positive fraction
We don't know status of n, insufficient.
Using both:
1/n  n <0> 0 > out
If n = positive integer, say 3, then 1/nn <0> possible
If n = negative fraction, say 1/3, then 1/nn <0> possible
If n = negative integer, say 3, then 1/nn > 0 > out
Two possibilites for n, insufficient.
E for me.



Director
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How would you rate this problem > Easy/Medium/Hard in terms of what you would find on the GMAT DS?



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Re: DS Modulus [#permalink]
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09 Jul 2007, 19:09
vshaunak@gmail.com wrote: Is n < 1 ? (1) (n^x) – n < 0 (2) x–1 = –2
(1) Insuf right away.
(2) Same as (1).
(1&2) x = 1, then: 1/n  n = (1n^2)/n < 1, which holds for (1, 0) and (1, oo+), therefore E.
Last edited by Andr359 on 09 Jul 2007, 19:53, edited 3 times in total.



Director
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vshaunak@gmail.com wrote: Think more. OA is not 'E'
now its your turn to release the OA.



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Re: DS Modulus [#permalink]
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09 Jul 2007, 22:06
vshaunak@gmail.com wrote: Is n < 1 ? (1) (n^x) – n < 0 (2) x–1 = –2
OA later...
I go with C.
From both statements together: (1n^2)/n < 0
If n < 1, both numerator and denominator will be less than 0, and (1n^2)/n will be greater than zero.
So, n can be within (1,0) to satisfy the inequality. Hence, n < 1. (C)



Director
Joined: 14 Jan 2007
Posts: 774

OA is 'C'.
Stmt1:
n^x n < 0
n[n^(x1)  1] < 0
n < 0 and n^(x1)  1 > 0 or n > 0 and n^(x1)  1 < 0
Lets say
n < 0 and n^(x1)  1 > 0
n = 4 , x =3 ===> n > 1
n=1/2, x =1 ====> n < 1
So INSUFF
Now n > 0 and n^(x1)  1 < 0
n =2 , x =5/4 ====>n > 1
n=1/2 , x=5 ====>n < 1
So INSUFF
Stmt2: clearly INSUFF
Taken them together:
n> 0 and n^(3/2) < 1 or n< 0 and n^(3/2) > 1
for n> 0,
n * sqrt(n) > 1 n > 1
For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.
Hence SUFF



Manager
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vshaunak@gmail.com wrote: OA is 'C'.
Stmt1: n^x n < 0 n[n^(x1)  1] < 0 n <0> 0 or n > 0 and n^(x1)  1 < 0
Lets say n <0> 0 n = 4 , x =3 ===> n > 1 n=1/2, x =1 ====> n <1> 0 and n^(x1)  1 <0>n > 1 n=1/2 , x=5 ====>n <1> 0 and n^(3/2) < 1 or n<0> 1 for n> 0, n * sqrt(n) > 1 n > 1
For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.
Hence SUFF
(1)n^x<n
(2)x=1
(1)+(2): n^(1)<n 1/n<n
case 1:if 0<n; 1<n^2; 1<n
case 2:if n<0; n^2<1; 1<n<0
Answer is E.
The two cases listed below will call answer C into question.
case 1: n=10, x=1; comply (1) and (2), 1<n
case 2: n=0.5, x=1; comply (1) and (2), n<1



Director
Joined: 16 May 2007
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I will go with C .
Agree with vshaunak.



Director
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wudy wrote: vshaunak@gmail.com wrote: OA is 'C'.
Stmt1: n^x n < 0 n[n^(x1)  1] < 0 n <0> 0 or n > 0 and n^(x1)  1 < 0
Lets say n <0> 0 n = 4 , x =3 ===> n > 1 n=1/2, x =1 ====> n <1> 0 and n^(x1)  1 <0>n > 1 n=1/2 , x=5 ====>n <1> 0 and n^(3/2) < 1 or n<0> 1 for n> 0, n * sqrt(n) > 1 n > 1
For n < 0 NOT POSSIBLE as sqrt(n) requires n to be +ve.
Hence SUFF (1)n^x<n (2)x=1 (1)+(2): n^(1)<n 1/n<n case 1:if 0<n; 1<n^2; 1<n case 2:if n<0; n^2<1; 1<n<0 Answer is E. The two cases listed below will call answer C into question. case 1: n=10, x=1; comply (1) and (2), 1<n case 2: n=0.5, x=1; comply (1) and (2), n<1
n can't be ve as n^3/2 will be undefined for ve values of n.



Director
Joined: 14 Jan 2007
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Apologies....
I put the stmt2 wrongly.
The correct stmt2 is x^1 = 2
I have edited the question.



Director
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vshaunak@gmail.com wrote: Apologies.... I put the stmt2 wrongly. The correct stmt2 is x^1 = 2 I have edited the question.



Director
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dahcrap wrote: vshaunak@gmail.com wrote: Apologies.... I put the stmt2 wrongly. The correct stmt2 is x^1 = 2 I have edited the question. Even still I would stick with E
seems it makes sense now cuz x = 1/2 and (n^x) <n> 1. if n is 1 or less than 1 but grater than 0, (n^x) < (n) doesnot hold true. so it should be C.



Senior Manager
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dahcrap wrote: vshaunak@gmail.com wrote: Apologies.... I put the stmt2 wrongly. The correct stmt2 is x^1 = 2 I have edited the question. Even still I would stick with E
I am not sure how to approach this problem... but
does n^1/2 mean we take the negative square root?
or do we go about it like this:
if n=1/4, (1/4)^1/2 = 1/sqrt(1/4) = 1/1/2 =2. 2.25>1
if n=4 4^1/2 = 1/sqrt4 = 1/2. .54 > 1
i'd go with C










