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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?

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

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, 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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03 Apr 2013, 12:36

1

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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.

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.

(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 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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