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If n is not equal to 0, is n < 4 ? [#permalink]
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08 Aug 2009, 06:55
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If n is not equal to 0, is n < 4 ? (1) n^2 > 16 (2) 1/n > n OPEN DISCUSSION OF THIS QUESTION IS HERE: ifnisnotequalto0isn41n85256.html
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Last edited by Bunuel on 17 Jul 2014, 01:30, edited 1 time in total.
Edited the question and added the OA.



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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08 Aug 2009, 07:24
lbsgmat wrote: If n is not equal to 0, is n < 4 ?
(1) n2 > 16
(2) 1/n > n D for me. 1) if n^2 > 16, then we get two possibilities, either n>4 or n<4. For both cases, absolute value of n will be always > 4. Sufficient. 2) In order for this inequality to be true, n can be anything < 1 (excluding 0 ). So we cant say if the absolute value is < 4. Insufficient.
Last edited by Economist on 08 Aug 2009, 10:02, edited 1 time in total.



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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08 Aug 2009, 07:30
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I would go for A
statement 1 is sufficient as n^2 > 16 meaning n > 4 or n < 4, either way, n > 4
statement 2 1/n > n...insufficient, two examples: when n = 2, 1/2 > 2, 2 < 4
when n = 5, 1/5 > 5, 5 > 4



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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16 Aug 2009, 00:47
8) If n is not equal to 0, is n < 4 ?
(1) n^2 > 16
(2) 1/n > n



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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16 Aug 2009, 03:07
Imo D given n!= 0, is n < 4? which is n^2 < 16? stmt 1 n^2>16 , so n^2 cannot be < 16. suffi.
stmt2 1/n > n squaring on both sides 1/(n^2) > n^2 1 > n^4 ( can multiply because n^2 is always +ve independent of the value of n) n^4<1, then defintely n^2 < 1, so n^2 < 16 suffi.



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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16 Aug 2009, 04:59
Yep D.
stmt 1: n>4 or n<4. In either case the magnitude of n (absolute value) will always be >4. Sufficient.
stmt 2: 1>n*n, For this inequality to be true, n<1. Hence, n will always be <4. Sufficient.
Last edited by Economist on 16 Aug 2009, 08:05, edited 1 time in total.



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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16 Aug 2009, 07:59
Economist wrote: stmt 2: 1>n*n, For this inequality to be true, we have 1<n<0 or 0<n<1. Hence, n will always be <4. Sufficient. I agree with D, but the solution for statement 2 is not \(1<n<0\) or \(0<n<1\) \(1<n<0\) or \(0<n<1\) is the solution for \(n^2 < 1\) Solution for \(n*n < 1\) is \(n < 1\)
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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 06:49
yezz wrote: crejoc wrote: Imo D given n!= 0, is n < 4? which is n^2 < 16? stmt 1 n^2>16 , so n^2 cannot be < 16. suffi.
stmt2 1/n > n squaring on both sides ( can we do that " what if n is ve , squaring will hide the sign??) 1/(n^2) > n^2 1 > n^4 ( can multiply because n^2 is always +ve independent of the value of n) n^4<1, then defintely n^2 < 1, so n^2 < 16 suffi. We cannot do that unless we are sure that n is +ve, else the inequality sign will reverse. Hence we can only derive 1>nn, as n is always +ve:)



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 11:34
Guys the answer is 'A'.
n*n < 1 ==> n < 0 it holds good for any negative values so insuff to say if n < 4 or not...



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 17:03
yezz wrote: 8) If n is not equal to 0, is n < 4 ?
(1) n^2 > 16
(2) 1/n > n Thats A. (1) If n^2 > 16, n is either > 4 or < 4. Suff.. (2) If 1/n > n, n is smaller than 1 but not 0 i.e. n could be 0.5, 1, 10 etc. not suff.. I was scrolling down for A. Got it............ skpMatcha wrote: Guys the answer is 'A'.
n*n < 1 ==> n < 0 it holds good for any negative values so insuff to say if n < 4 or not...
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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 17:05
Economist wrote: squaring on both sides ( can we do that " what if n is ve , squaring will hide the sign??) We cannot do that unless we are sure that n is +ve, else the inequality sign will reverse. Hence we can only derive 1>nn, as n is always +ve:) that makes sense economist,.. i thought squaring will hide it.. you are correct it cannot be squared, unless n is always +ve..



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 17:29
I dont understand, doesn't statement 2 yield N to be under 1 and statement 1 yields it to be above 4. Can this be a question, N cant be both? Either way, arent both statement sufficient, statement 1 tells you it has to be over 4, statement 2 tells you definitively that its under it?



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Aug 2009, 18:04
sfeiner wrote: I dont understand, doesn't statement 2 yield N to be under 1 and statement 1 yields it to be above 4. Can this be a question, N cant be both? Either way, arent both statement sufficient, statement 1 tells you it has to be over 4, statement 2 tells you definitively that its under it? i think stmt1 is clear and is sufficient, and you have no doubt with it. consider stmt2 question is n< 4? stmt2 given 1/n>n case 1: take when n is +ve 1 > n* n 1> n^2 n^2<1 so 1<n<1 in this interval for any values n is < 4 case 2: when n is ve 1/n> n 1/(n)>n multiply by n 1< n^2 n^2>1 n^2>1 since n is negative, and n^2>1, for all negative values of n this is true, so n can take any negative values say 1,2..and so on if n=2 1/n>n, 1/2> 2 is true, check n < 4, 2<4, 2< 4 true, if n= 8 1/n>n, 1/8 > 8 , but when you check n<4, 8<4? , no 8 is not < 4, so insufficient Another easy way is to directly check by plugging numbers stmt2 1/n> n for any positive value of n, this statement holds false, so n can take negative values and only fractions. check with negative numbers when n = 2 n < 4? , 2<4 , yes but when n = 8 n < 4?, 8is not < 4, so stmt2 insufficient similarly, for fractions also try using n = 1/2, n=1/8 , it is insufficient so answer is A hope this helps..



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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15 Sep 2009, 02:16
Agree with A. It is the sole choice, which represents a clear answer on the question. In stmt B there is an evidence, that n is negative only. This gives nothing.
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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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23 Dec 2009, 17:21
Can somebody breakdown for me how n^2 > 16 is n<4 or n>4.
As I look at it if n^2 > 16 then n > + or 4.
also I think I understand how n > 4 but please breakdown that as well.
Thanks.



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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23 Dec 2009, 18:52
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sudimba wrote: Can somebody breakdown for me how n^2 > 16 is n<4 or n>4.
As I look at it if n^2 > 16 then n > + or 4.
also I think I understand how n > 4 but please breakdown that as well.
Thanks. n^2>16>+n>16>+n>16 or n>16>n<4(multiplying bth side by 1 reverses the sign) one can try with numbers also as square of(3)=9<16 therfore n<4 to satisfy the inequality same with otherone lnl>4>+n>4 or n > 4>n<4
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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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24 Dec 2009, 16:38
From (i)
\(n^2>16\) \(\sqrt{n^2}>\sqrt{16}\) \(n>\sqrt{16}\) \(n>4\)
/* Because \(n = \sqrt{n^2}\) applies only for positive square roots of n */ Therefore, n<4 is not TRUE.
From(i) we could find this answer.
From(ii)
\(1/n>n\)
\(nn<1\) \(n<1/n\) If n>0, \(n<1/n\) \(n^2<1\) \(n<1\) n<4 is TRUE.
or If n<0, \(n<1/n\) \(]n>1/n\) \(n^2>1\) \(n>1\) n<4 may or may not be TRUE.
Therefore from(ii), we can't answer this question.
Thus answer is "A"



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Re: If n is not equal to 0, is n < 4 ? [#permalink]
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17 Jul 2014, 01:30
If n is not equal to 0, is n < 4 ? Question basically asks is 4<n<4 true. (1) n^2>16 > n>4 or n<4, the answer to the question is NO. Sufficient. (2) 1/n > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient. Answer: A. As you can see we don't really want the complete range for (2) to see that this statement is not sufficient, but still if interested: 1/n > n > n*n < 1. If n<0, then we'll have n^2<1 > n^2>1. Which is true. So, n*n < 1 holds true for any negative value of n. If n>0, then we'll have n^2<1 > 1<n<1. So, n*n < 1 also holds true for 0<n<1. Thus 1/n > n holds true if n<0 and 0<n<1. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifnisnotequalto0isn41n85256.html
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