If n is not equal to 0, is |n| < 4 ?

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

8) If n is not equal to 0, is |n| < 4 ?

(1) n^2 > 16

(2) 1/|n| > n

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.

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.

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 \(n < 1\)

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:)

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...

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............

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 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.

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

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)

\(1/|n|>|n|\)

\(n|n|<1\)
\(|n|<1/n\)
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"