If x^2 < x and x is written as a terminating decimal, does x have a nonzero hundredths digit?

(1) 10x is not an integer.
(2) 100x is an integer.

First, translate the given information: if x^2 < x, that's code for 0 < x < 1, because only numbers in that range will have squares that are smaller than themselves.

So, if x is between 0 and 1, and it's written as a terminating decimal, does it have a nonzero hundredths digit? In other words, does the decimal continue to at least the hundredths place?

1) 10x is not an integer.

This tells us that multiplying by 10 is not enough to make this a whole number. In other words, the decimal does not end after the tenths place (e.g. it can't be 0.4). You might think that if the decimal continues, there must be a non-zero hundredths place. But try to prove insufficiency!

ex: x = 0.412 --> this obeys the statement, and would give us a "yes" answer to the question
ex: x = 0.402 --> this obeys the statement, but it would give us a "NO" answer to the question

Insufficient.

2) 100x is an integer.

This tells us that multiplying the number by 100 is enough to make it a whole number. Thus, the decimal could not continue after the hundredths place (e.g. it could not be 0.456). But is that enough to tell us whether there is a non-zero hundredths place?

ex: x = 0.41 --> this obeys the statement, and would give us a "yes" answer to the question
ex: x = 0.4 --> this obeys the statement, but it would give us a "NO" answer to the question

Insufficient.

1 & 2) 10x is not an integer but 100x is an integer.

This tells us that x must have exactly 2 non-zero digits after the decimal place, since those are the only numbers that would be consistent with both statements.

Sufficient. The answer is C.