If n is an integer, is n – 1 > 0? : GMAT Data Sufficiency (DS)
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# If n is an integer, is n – 1 > 0?

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If n is an integer, is n – 1 > 0? [#permalink]

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24 Apr 2012, 21:31
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If n is an integer, is n – 1 > 0?

(1) n^2 – n > 0

(2) n^2 = 9

[Reveal] Spoiler: OA
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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24 Apr 2012, 21:37
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If n is an integer is n - 1 > 0

Is n-1>0? --> is n>1?

(1) n^2 - n > 0 --> n(n-1)>0 --> n<0 or n>1. Not sufficient.

(2) n^2 = 9 --> n=3 or n=-3. Not sufficient.

(1)+(2) n could still be 3 or -3. Not sufficient.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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25 Apr 2012, 00:20
None of the option alone , together are sufficient to determine whether n>1
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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25 Apr 2012, 03:49
1) n^2-n>0 : in this case if n is -ve thn this case is always true, but the eqn n-1>0 becomes invalid, hence insufficient.
2) n^2=9 => n=3 or n=-3, hence insufficient.
both are insufficient.
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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29 Apr 2012, 20:23
Thanks for the replies but I still have a doubt. Shouldn't the answer be C?
1) n^2 - n > 0 --> n(n-1)>0, doesn't this mean that n>0 or n>1 so when we combine this with n= +3 or n= -3 then the answer is +3?

Thank you!
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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29 Apr 2012, 22:19
rakp wrote:
Thanks for the replies but I still have a doubt. Shouldn't the answer be C?
1) n^2 - n > 0 --> n(n-1)>0, doesn't this mean that n>0 or n>1 so when we combine this with n= +3 or n= -3 then the answer is +3?

Thank you!

The range n>0 or n>1 does not make any sense.

(1) n^2 - n > 0 --> n(n-1)>0 --> the roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root: n<0 or n>1. Not sufficient.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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29 Apr 2012, 22:28
Thank you for the clarification. I get it now!
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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29 Apr 2012, 23:34
Ans is C
(1) n^2 - n > 0 , so n > 0 or n - 1>0 ----> insufficient
(2) n^2 = 9, so n= 3 or n = -3 -----> insufficient
Combine: we have n=3, hence n -1 . 0
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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29 Apr 2012, 23:37
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carriehoang0410 wrote:
Ans is C
(1) n^2 - n > 0 , so n > 0 or n - 1>0 ----> insufficient
(2) n^2 = 9, so n= 3 or n = -3 -----> insufficient
Combine: we have n=3, hence n -1 . 0

The red part is not correct. Refer to my previous post.
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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03 Apr 2013, 09:24
Bunuel could you please illustrate this concept thoroughly. I am having trouble understanding this one. n^2 - n > 0 --> n(n-1)>0 --> n<0 or n>1. Normally, we take it as n>0 and n>1. Needed extra light...
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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03 Apr 2013, 11:36
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Hello atalpanditgmat,

Here are my two cents on this.

In case of n(n-1)>0, we need to find solutions so that the product of n and n-1 is positive. This is possible only if both the values are positive or negative.
This can happen under two conditions.
n>1 so that n and n-1 are positive
n<0 so that n and n-1 are negative.

Now, discussing your doubt in details, suppose the equation to be x(y-1)>0. In such an equation, x >0 and y>1, so that the product of x and y-1 is positive. The other possibility is x<0 and y<1 so that the product of two negatives turns out positive. When, the same number is being considered, we cannot consider the variable in each of the expressions(like n and (n-1)) independent of each other as in case of x and y.

Hope this helps.

atalpanditgmat wrote:
Bunuel could you please illustrate this concept thoroughly. I am having trouble understanding this one. n^2 - n > 0 --> n(n-1)>0 --> n<0 or n>1. Normally, we take it as n>0 and n>1. Needed extra light...
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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04 Apr 2013, 03:07
atalpanditgmat wrote:
Bunuel could you please illustrate this concept thoroughly. I am having trouble understanding this one. n^2 - n > 0 --> n(n-1)>0 --> n<0 or n>1. Normally, we take it as n>0 and n>1. Needed extra light...

Check here: if-n-is-an-integer-is-n-131293.html#p1080057

Hope it helps.
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Re: If n is an integer, is n – 1 > 0? [#permalink]

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04 Apr 2013, 05:41
Thanks Kris01, Your explanation helped a lot.....
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Re: If n is an integer, is n – 1 > 0?   [#permalink] 04 Apr 2013, 05:41
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