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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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rakp
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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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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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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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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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
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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atalpanditgmat
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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Bunuel
rxs0005
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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Bunuel
rxs0005
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:
Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

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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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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If n is an integer is n - 1 > 0

(1) n^2 - n > 0 --> n(n-1) > 0
n>2 WORKS
n<0 DOES NOT
Insufficient.

(2) n^2 = 9
|n| = 3 --> n = 3,-3
Insufficient

Combo pack:
Nothing has changed from st2 even after accounting for st1. n = 3,-3
Insufficient.
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