Is square root of x*y an integer? 1). Root x is not an

Question

Is square root of x*y an integer?
1). Root x is not an integer
2). Root y is not an integer

eskay05 · Answer

For sqrt(xy) to be an integer:

xy must be an integer AND
either xy must be a perfect square or x and y must each be perfect squares.

(1) and (2) don't give all this information. 

Hence answer = E

I'm not sure about my explanation though .. does anyone agree?

Thanks

pmenon · Answer

I would say D.

sqrt(x*y) = sqrt(x)*sqrt(y)

From statement 1, we know that root x is not an integer. Therefore, no matter what sqrt(y) turns out to be, when it gets multiplied by the value of sqrt(x), the product will not be an integer in any case.

Same logic for statement 2.

Does anyone see a flaw in this logic ? Thanks !

vivek123 · Answer

Your logic will not be valid, since we don't know about "sqrt(y)". It can be an integer or non-integer that will multiply non-integer "sqrt(x)" & make result an integer or non-integer! Say for example...

what if, 
sqrt(x) = 1.5 & sqrt(y)=2 ?  YES, result is integer 
AND
sqrt(x) = 1.5 & sqrt(y)=1.5 ?  NO, result is not an integer 


It should be 'E'!

pmenon · Answer

Vivek, in your first example, sqrt of (1.5 x 2) is still not an integer. What I am saying is that if we know already that ONE of the roots is not an integer, it wont matter if the other root comes out to an integer or not, since the root of the PRODUCT wont be an integer, and thats what the question is asking. Right ?

pmenon · Answer

let me give an example. say x=3 and y=4

sqrt(xy)=sqrt(12)=sqrt(3)*sqrt(4).

Sqrt(4) is an integer, but sqrt(3) is not. Therefore, there is no way that sqrt(xy) will be an integer itself.

This comes from statement 1 when we are told that root x is not an integer.

I hope this makes sense, I really dont see it any other way

vivek123 · Answer

pmenon, I think you are mistaking it... 
You are taking sqrt of (1.5 x 2) again?

I said, sqrt(x) = 1.5 & sqrt(y)=2
& 
sqrt(x*y) = sqrt(x)*sqrt(y) = 1.5x2 = 3.
is this correct?

pmenon · Answer

AAAAAAAAAAHHH Sorry for the confusion folks. Vivek, youre right, for some reason i was taking the root of 3 again !!! I change my answer to E 8-)

Professor · Answer

should be E.
from i, if x = 3 and y = 2, sqrt(xy) = is not an integer.
if x = 3 and y = 3, sqrt(xy) = is an integer.

from ii, if y = 3 and x = 2, sqrt(xy) = is not an integer.
if y = 3 and x = 3, sqrt(xy) = is an integer.

from i and ii, if x = 2, y = 3, no. if x=y=3, yes. so insuff.

Is square root of x*y an integer? 1). Root x is not an

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards