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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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16 Feb 2011, 05:57
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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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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:
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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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, 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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04 Mar 2014, 10:03
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.

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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04 Mar 2014, 10:10
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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.

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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04 Mar 2014, 20:25
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.

If this was helpful, any kudos would be appreciated.
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Re: If n is an integer is n - 1 > 0? [#permalink]

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