II wrote:From this we can say that if n^2 < 1/100, then n < 1/10, or n < -(1/10)

lunarpower wrote:one mistake in here that deserves attention:

II wrote:From this we can say that if n^2 < 1/100, then n < 1/10, or n < -(1/10)

whoa, no.

if you're given that n^2 is less than a given value, then n must be less than the positive square root (as you've indicated), but greater than the negative square root.

so, in this case, we have -1/10 < n < 1/10. if you write this out as two separate inequalities (as you did here), then those two inequalities should be n < 1/10 and n > -1/10.

because n has to be negative, then, n itself must be between -1/10 and 0. this means that the reciprocal of n is between negative infinity and -10.

you got lucky here, because there's only one answer choice that even has -10 as a boundary value. notice that you managed to solve the problem without even thinking about whether your reciprocal must be greater or less than -10; it was good enough just to look for the answer with -10 in it.
if there were confounding answer choices, such as 'between -10 and 0', you'd have to think about whether the reciprocal itself is less/greater than -10.

II wrote:Hi Lunarpower ... sorry ... dont think I understand your comments. Can you please elaborate.

n^2 < 1/100
n < 1/10
reciprocal of 1/10 is 10.
so n < 10
n is a negative number ... so we get the -10.
n < -10.

How would you solve this question ?

lunarpower wrote: one thing that appears to be going wrong here is that you're taking reciprocals, but NOT flipping the inequality sign around. that's bad.
for instance, if x < 1/2, then 1/x > 2.