n^2<1/100, follows that -1/10<n<0
n>-1/10
(n)^(-1)<(-1/10)^(-1), n^(-1)=1/n, (-1/10)^(-1)=-10. we reverse the sign of the inequality since for example if 2<3, then 1/2>1/3, or -3<-2, then -1/3>-1/2 so
1/n<-10

n^2<1/100, follows that -1/10<n<0.....how come we are coming to the result that -1/10 < n < 0

if n^2<1/100. n^2-(1/10)^2<0
(n-1/10)(n+1/10)<0, this inequality is valid for -1/10<n<1/10. but as n<0. the resulting is
-1/10<n<0

Given: 1. n is on left side of 0 on number line. This simply means n is negative.
2. n^2 < 1/100

Taking square root on both sides
(n^2)^(1/2) < (1/100)^(1/2)
|n| < 1/10
-1/10 < n < 1/10
But as n is negative, -1/10 < n < 0
so, n> -1/10
Multiplying -1 both sides
-n < 1/10 ( sign changes as -ve number is multiplied)
Multiplying 10 on both sides
-10n < 1 ( sign does not change as multiplied by 10 (+ve) )
Dividing n on both sides
-10 > 1/n ( Again sign changes as n is -ve)

Answer is A.