Is x > 10^10 ?
(1) x > 2^34
(2) x = 2^35
OA: D
EDIT: please note, both statements have been changed to include the exponent sign, as they had not been transfered in the original copy/paste.
Is x>10^10
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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
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
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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)
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)
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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.
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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