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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.

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>|n|n, as |n| is always +ve:)

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>|n|n, 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..

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 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?, |-8|is 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..

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]
23 Dec 2009, 17:52

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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