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If x is a positive integer is sqrt(x) an integer? 1) sqrt(4x) is an integer 2) sqrt(3x) is an integer [I'm not sure if the OA is correct]

from 1) sqrt(4x) = sqrt(4)*sqrt(x) so 2*sqrt(x) is an integer sqrt can be 3/2 of 1/2 so is insuff from 2) same stuff sqrt(3x) = sqrt(3)*sqrt(x) so sqrt(3)*sqrt(x) is an integer so sqrt(x) is not an integer so i would go with B

well the OA is A. However, i don't think that is correct and here is my reasoning:

1.sqrt(4x) is an integer -> so 2 time sqrt(x) is an integer --> but sqrt(x) could be = 1.5, 2.5 3.5 or any such number so that 2sqrt(x) = 3, 5, 7 etc.
therefore just because 2sqrt(x) is an integer, x does not have to be an integer. Not Suff!

2. sqrt(3x) is an integer --> i.e. 3x is a perfect square but x could be 4/3, which would make sqrt(3x)=sqrt(4)=2 an integer. On the other hand, x could be 3 which would make sqrt(3x) = 3 an integer as well. Not Suff!

1 & 2 together: 4x and 3x is a perfect square. I can't think of any number where both these conditions are true.

Am I missing something??

PS. This question is from the OG Quant Review, DS #116

If x is a positive integer is sqrt(x) an integer? 1) sqrt(4x) is an integer 2) sqrt(3x) is an integer

[I'm not sure if the OA is correct]

X > 0, integer

(1) sqrt(4x) is an integer => sqrt 4 * sqrt x = integer = > sqrt x has to be an integer. Sufficient.

(2) sqrt(3x) is not an integer => sqrt 3 * sqrt x => sqrt 3 is obviously not an integer. If x = 1, sqrt 3 is not an integer, but sqrt 1 is. Insufficient

A ) sqrt(4x) = Integer take no 16,36....sqrt of these no is Integer and sqrt(x) integer ...Suff

B)
sqrt(3x) = Integer, sprt(144) = sqrt(3*48)=Integer but sqrt(48) is not integer sqrt(81) = sqrt(3*27)=Integer but sqrt(27) not Integer
sqrt(36) = sqrt(3*12)=Integer but sqrt(12) not integer... always NO so SUFF...

A ) sqrt(4x) = Integer take no 16,36....sqrt of these no is Integer and sqrt(x) integer ...Suff

B) sqrt(3x) = Integer, sprt(144) = sqrt(3*48)=Integer but sqrt(48) is not integer sqrt(81) = sqrt(3*27)=Integer but sqrt(27) not Integer sqrt(36) = sqrt(3*12)=Integer but sqrt(12) not integer... always NO so SUFF...

both are sufficient so ans D

cool_jonny009,

with statement 2

you can use x=4
sqrt(3*4) is not an integer
but sqrt(4) is an integer

you can also use x=2
sqrt(3*2) is not an integer
and sqrt(2) is not an integer