acerche wrote:If x is a positive integer, is x^0.5 an integer?

1) 4x^0.5 is an integer

2) 3X^0.5 is not an integer

ANS: A

Can someone CLEARLY explain this. Book explanation is TERRIBLE!

Step 1 of the Kaplan method for DS: focus on the question stem.

x^(.5) is the root of x. So, what the question is really asking us is:

If x is an integer, is x a perfect square?

Step 2 of the Kaplan method: examine each statement, by itself.

(1) (4x)^.5 is an integer
[I'm assuming that's where the bracket goes]

Well, we can rewrite (4x)^.5 as (4)^.5 * (x)^.5

(4)^.5 * (x)^.5 = 2*rootx

Well, given that x is an integer, the only way for 2rootx to be an integer is if rootx is an integer. If rootx is an integer, then x is a perfect square: sufficient.

(note: it could also work out to -2*-rootx, but for the purposes of whether x is an integer, sign is irrelevant)

(2) (3x)^0.5 is not an integer

In other words, 3x is NOT a perfect square.

Well, it's hard to do mathematical proofs for "NOT" statements, so let's pick numbers to show we can get a "yes" and a "no":

if we pick x = 1, then 3x = 3 is NOT a perfect square.
Is 1 a perfect square? YES

if we pick x = 2, then 3x = 6 is NOT a perfect square.
Is 2 a perfect square? NO

So, statement (2) can give us both a "yes" and a "no" answer: insufficient.

(1) is suff, (2) is insuff: choose (A).

Thanks for the great explanation!