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

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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.

Answer: E.


Solving inequalities:
http://gmatclub.com/forum/x2-4x-94661.html#p731476 (check this one first)
http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... me#p873535
http://gmatclub.com/forum/everything-is ... me#p868863

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

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New post 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.

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

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New post 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?

Please elaborate.
Thank you!

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

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New post 29 Apr 2012, 23:19
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?

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.

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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New post 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

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

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

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

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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.

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

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New post 04 Apr 2013, 04:07
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...
Thanks in advance.


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

Hope it helps.
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New post 04 Mar 2014, 11:03
Bunuel wrote:
rxs0005 wrote:
If n is an integer is n - 1 > 0


S1 n^2 - n > 0

S2 n^2 = 9


If n is an integer is n - 1 > 0

Question: 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.

Answer: E.


Hello Bunuel

Please Clarify how did you get : \(n<0\) or \(n>1\). from \(n(n-1)>0\)

If i Try
frm eq : \(n(n-1)>0\)
We can say that

either both are positive : \(n>0\) or \((n-1)>0\) -> \(n>1\) ... SO probably on combining these 2 equations we will get : \(n>1\)

or both are negative : \(n<0\) or \((n-1)<0\) -> \(n<1\) .. here on combining we can get : \(n<1\)

Please tell me where am i wrong ?

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New post 04 Mar 2014, 11:10
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niyantg wrote:
Bunuel wrote:
rxs0005 wrote:
If n is an integer is n - 1 > 0


S1 n^2 - n > 0

S2 n^2 = 9


If n is an integer is n - 1 > 0

Question: 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.

Answer: E.


Hello Bunuel

Please Clarify how did you get : \(n<0\) or \(n>1\). from \(n(n-1)>0\)

If i Try
frm eq : \(n(n-1)>0\)
We can say that

either both are positive : \(n>0\) or \((n-1)>0\) -> \(n>1\) ... SO probably on combining these 2 equations we will get : \(n>1\)

or both are negative : \(n<0\) or \((n-1)<0\) -> \(n<1\) .. here on combining we can get : \(n<1\)

Please tell me where am i wrong ?


\(n(n-1)>0\) means that the multiples have the same sign:

n>0 and n-1>0 --> n>0 and n>1 --> n>1 (both to hold true n must be greater than 1);
n<0 and n-1<0 --> n<0 and n<1 --> n<0 (both to hold true n must be less than 0).

There are other approaches possible which are explained in the posts below:


Also check our new set, 50 GMAT Data Sufficiency Questions on Inequalities: 50-gmat-data-sufficiency-questions-on-inequalities-167841.html

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

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New post 04 Mar 2014, 21:25
Also an alternative solution to the problem is the following:

Statement 1:

n^2-n>0 --> n^2>n (you can always use addition in inequalities) --> 1>1/n (we can divide by n^2 since we know that it is positive since any number squared is positive).

So from this statement we know that 1/n is less than 1. That means that n>1, since 1 divided by a number greater than 1 results in a fraction or n is negative. This is not enough to answer the question

Statement 2:

This statement tells us that n is -3 or 3. Not enough to answer the question either. When we combine both statements we still can not answer, therefore making the answer E.

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