Is x>10^10

IMO B.

Stmt 1. x > 254

X can be any number from 255 to infinity. Depending on the value, x could be < 10^10 OR = 10^10 OR > 10^10. So insufficient.
(any number < 10 ^10 could have been used in stmt 1 (

Stmt 2.
x = 255.

So x definitely < 10^20.
Always false. So sufficient.

Hence B

sorry vittal - I forgot to include the exponents in the question.

Try again.

Stmt II

No need to do anyhting but say SUFF since we have x=2^35 which is one definite value from which we can say if its > 10^10 or not

Stmt I

x>2^34 i.e x> 2^10 * 2^24 (2^2 can be approximated to 5^1)

10^10 = 2^10 * 5^10


x> 2^10 * 5 ^ 12

SUFF

D)

The answer also had an "approximation". For some reason, I don't feel comfortable doing that. I guess there really isn't any other way though.

Thanks!

I guess there really isn't any other way though

IMO Yeah approximation would be the way to go!

3^3 = 5^2

2^2= 5^1

first thing to do is break 10^10 into something easier to work with:

10^10 = (5x2)^10 = 5^10 x 2^10

Can't really do much more with it than that.

For the 1st equation, x > 2^34 we know that if 2^34 > 10^10, then this answer is sufficient. If not, x can potentially be a number lower than 10>10, but also higher because we know only that it is greater than 2^34, but that could mean any number onward.

so we break 2^34 down to something easier to work with:

2^34 = 2^24 x 2^10 = 4^12 x 2^10

at this stage, all you need to do is figure out whether 4^12 > 5^10

I (correctly) guessed that it was, but if someone has a better way to figure out that it is, I'd love to hear it.

Same procedure for the second one if you care to solve for x, but we have a fixed value, so we really don't need to. you can for fun I guess, but there's no need, you know (2) is sufficient.

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