Inequality

[kutra]

If n is not equal to 0, is |n| < 4 (Does this mean n² < 16 ) ?

(1) n² > 16
(2) 1/|n| > n

Hiow would you solve this algebraically?
I seem to be getting IMO as D when I do it the squaring both sides method for stmt 2?

1 > n|n|
Squaring both sides
1 > n²n²
n⁴ < 1




HELP!

[finsisher]

imo D

ist stmt, n^2>16 implies n >4 or n<-4 sufficient

2nds stmt 1 > n mod n ( since mod n is positive we can multiply both sides by mod n without changing the sign of inequality)
implies n^2 <1 implies n is b/w -1 and 1

hence D

[sh_vivek]

Is something wrong with the question?

From I: n<-4 or n>4
From II:-1<n<1

The range of n from I and II does not overlap. IN GMAT DS, both statements do not contradict each other.

[ForestGump]

IMO A!!
(i) it implies n >4 and n<-4...in both the cases mod n is >4

whereas
(ii) n!n! < 1..
!n! will always b +ve..
n can have any value..fractions ,-ve, but no +ve value greater than 1.
so we n

[ForestGump]

IMO A!!
(i) it implies n >4 and n<-4...in both the cases mod n is >4

whereas
(ii) n!n! < 1..
!n! will always b +ve..
n can have any value..fractions ,-ve, but no +ve value greater than 1.
so we are not sure abt the value of !n!..

[Fiver]

Originally Posted by sh_vivek

Is something wrong with the question?

From I: n<-4 or n>4
From II:-1<n<1

The range of n from I and II does not overlap. IN GMAT DS, both statements do not contradict each other.

Actually the stmts do not contradict each other.
Stmt1] means |n| > 16. Therefore n>4 or n<-4
Stmt2] means that n is either a proper fraction or a negative number, which means that n<-4 is still applicable; hence the answer could be a no or a yes respectively.
My answer is A.
What's the OA?

[kutra]

OA is A. What I fail to understand is, if Stmt 2 is done by squariing both sides, OA should be D.

[clock60]

i think that the answer A
is -4<x<4-???
(1) n^2-16>0
(n-4)(n+4)>0 the given expression is valid for x>4 and x<-4 and the answer is definitely no so sufficient
(2) the second is tricky one
i am not sure that we can square as x may be negative
so two cases exists x>0 and x<0
if x>0
1/|x|=1/x and 1/x>x, (1-x^2)>0, (1-x)(1+x)>0 x=1 and x=-1
valid for interval (-1,1) but x>0 so (0,1)
if x<0
-(1/x)-x>0
1/x+x<0
(1+x^2)/x<0
1+x^2>0 in all cases
so x<0
and it follows for
1/|x|>x to be valid x must be less than 1 and not equal to 0
let us check for different values of x
if x=0,5, 1/0,5=2, 2>0,5
if x=-0,5 1/0,5=2 2>-0,5
if x=2 1/2<2
from second st we know that x<4 ( as is it less that 1) but it can be greater or lesser than -4
so A