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

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25 Apr 2012, 04:49

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.

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

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29 Apr 2012, 21:23

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?

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

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29 Apr 2012, 23:19

Expert's post

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?

Please elaborate. 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.

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

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30 Apr 2012, 00:34

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

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

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30 Apr 2012, 00:37

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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. _________________

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

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03 Apr 2013, 10:24

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... Thanks in advance. _________________

Do not forget to hit the Kudos button on your left if you find my post helpful.

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

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03 Apr 2013, 12:36

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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... Thanks in advance.

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

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04 Apr 2013, 04:07

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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... Thanks in advance.

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