Carcass, a little confused here. Could you please help?

n<10 as given in the question.
as n^2 < 1/100
Taking square root on either side, n< 1/10 and n>-1/10
so, -1/10 <n< 1/10

How to arrive at the answer after this?

We have \(n<0\) and \(n^2<\frac{1}{100}\)

\(n^2<\frac{1}{100}\) --> \(-\frac{1}{10}<n<\frac{1}{10}\), but as \(n<0\) --> \(-\frac{1}{10}<n<0\).

Multiply the inequality by \(-\frac{10}{n}\), (note as \(n<0\), then \(-\frac{10}{n}>0\), and we don't have to switch signs) --> \((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\) --> so finally we'll get \(\frac{1}{n}<-10<0\).

Answer: A.

Carcass thank you for the prompt reply. I always tend to get confused and ultimately mess up when multiplication/ division has to be done with a negative number in inequalities.
So, thank you for the response and kudos!