Thanks!

[Just Now]

Thanks!

[beatthegmatinsept]

Wrong place. This thread should be in the Data Sufficiency sub-forum.
I don't have OG12 handy with me, so if you don't mind posting the problem here, I can help you out.

[Just Now]

Sorry for posting in the wrong place (just started here). But thank you in advance for helping! Here's the problem: If n is a positive integer, is (1/10)^n <0.01?
(1) n>2
(2) (1/10)^n-1 < 0.1

I thought only A would be sufficient. But turns out either statement is.

[InkyBinky]

Sorry for posting in the wrong place (just started here). But thank you in advance for helping! Here's the problem: If n is a positive integer, is (1/10)^n <0.01?
(1) n>2
(2) (1/10)^n-1 < 0.1

I thought only A would be sufficient. But turns out either statement is.

Yeah, next time put it in the math section, but I'll answer it here.

Per my response to your general question on DS, put the given equation into a more useful form:
(1/10)^n = 10^-n
0.01 = 10^-2
So...
10^-n < 10^-2
Solving for n, the question is really asking: Is n > 2?

Knowing this, part 1 is obviously sufficient. Now look at part 2 and manipulate:

(1/10)^(n-1) < 10^-1

(1/10)^(n-1) = [(1/10)^n] * (1/10)^-1 = (10^-n) * 10

Dividing each side by 10, you get:

10^-n < 10^-2

This is a restatement of the second to last step in the original problem's manipulation above, so it answers the question and is sufficient.

[anantbhatia]

(1) itself is sufficient

(2) is (1/10)^n-1<0.1
so (1/10)^n<0.01 ie. LHS < RHS. That's what we wanted to know... Hence (2) being an assertion, answers the question if LHS was actually < RHS.

Is that understanding correct?