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Roots question

by sachin_yadav » Fri Apr 01, 2011 8:07 am
If x is a positive integer, is sqroot x an integer ?

(1). sqroot 4x is an integer. (both 4 and x are under square root)
(2). sqroot 3x is not an integer. (both 3 and x are under square root)

Answer is A

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Sachin
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by kmittal82 » Fri Apr 01, 2011 8:15 am
Don't see why its A

(1)
sqrt(4x) = 2sqrt(x) [2 with a + or -]

Now, for the above to be an integer, sqrt(x) could be 1.5, 2.5 etc, or sqrt(x) could be 3
So, its shouldn't be sufficient.

(2)
sqrt(3) x sqrt(x).
This can only be an integer if x = 3. For all other values, its not an integer, so the question is basically saying x is not equal to 3

Combine (1) + (2), all we know is x is not 3, so still not sufficient to answer the question

Should be (E) I think.
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by ldoolitt » Fri Apr 01, 2011 1:05 pm
kmittal82 wrote:Don't see why its A

(1)
sqrt(4x) = 2sqrt(x) [2 with a + or -]

Now, for the above to be an integer, sqrt(x) could be 1.5, 2.5 etc, or sqrt(x) could be 3
So, its shouldn't be sufficient.

(2)
sqrt(3) x sqrt(x).
This can only be an integer if x = 3. For all other values, its not an integer, so the question is basically saying x is not equal to 3

Combine (1) + (2), all we know is x is not 3, so still not sufficient to answer the question

Should be (E) I think.
I just wanted to point out the flawed reasoning. sqrt(x) cannot be 1.5 or 2.5 because 1.5 and 2.5 squared are not integers and the problem stem specifically states that x must be an integer.

The rule that you are looking for here is the following:

"The square root of an integer that is not a perfect square is irrational"

And the product of an irrational number and 2 is always irrational. Hence x must be a perfect square to satisfy statement (1) and by definition sqrt(x) is an integer.
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by atulmangal » Sat Apr 02, 2011 6:39 am
Can also see in this way...for Op A

sqroot 4x = 2*sqroot x

Now, 2*sqroot x = INT only if

condition 1:- sqroot x = vales like 1.5, 2.5 or y.5 (decimal value = 5)

NO Int value of X yields the above value

condition 2:- sqroot x = INT

Now condition 1 is INVALID as given the question that X = +ve INT

so it means sqroot x = INT

Hence A
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by kmittal82 » Sat Apr 02, 2011 3:21 pm
Yikes, I'm making too many mistakes, thanks for pointing out my flawed reasoning :)
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by ldoolitt » Sat Apr 02, 2011 6:19 pm
kmittal82 wrote:Yikes, I'm making too many mistakes, thanks for pointing out my flawed reasoning :)
Ha I didn't realize that came out sort of condescending (the flawed reasoning part). No harm meant!
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by force5 » Sun Apr 03, 2011 5:48 am
IMO A
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by sachin_yadav » Sun Apr 03, 2011 8:43 pm
Thank you very much ldoolitt and atulmangal. Really appreciate for your help.

Thanks & Regards
Sachin
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