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From (2) X and Y could be either positive or negative, thus, (2) alone is also not sufficient.

Combine => X&Y are positive.

Looks like the answer is E.

IMO, I think you need to re=examine your analysis for condition (1).
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Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Through the discussion people agreed that the answer is A, since on the GMAT we are dealing only with arithmetic radicals and, therefore, sqrt(X) and sqrt(Y) are non negative.

Disagree on that answer is:

USEFUL TO KNOW A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Is x>y?

(1) \(\sqrt{x}>\sqrt{y}\) --> as both parts of the inequality are non-negative then according to A we can square them --> \(x>y\). Sufficient.

(2) \(x^2>y^2\) --> \(|x|>|y|\) --> \(x\) is farther from zero than \(y\), but this info is insufficient to say whether \(x>y\) (if \(x=2\) and \(y=1\) - YES but \(x=-2\) and \(y=1\) - NO). Not sufficient.

1, square root X > square root Y Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD... 2, X^2>Y^2 here x= 1,y= 2 so 1 is not greater than 2, so

1, square root X > square root Y Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD... 2, X^2>Y^2 here x= 1,y= 2 so 1 is not greater than 2, so

only A....

can we consider negative values in square root??

You cannot get sufficiency based only on one set of numbers. Thus theoretically testing only one set of numbers can give you an incorrect answer.

As for your other questions: the even roots from negative numbers are not defined for the GMAT, which means that for \(\sqrt{x}\) to be defined, x must be more than or equal to zero.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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1, square root X > square root Y Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD... 2, X^2>Y^2 here x= 1,y= 2 so 1 is not greater than 2, so

only A....

can we consider negative values in square root??

You cannot get sufficiency based only on one set of numbers. Thus theoretically testing only one set of numbers can give you an incorrect answer.

As for your other questions: the even roots from negative numbers are not defined for the GMAT, which means that for \(\sqrt{x}\) to be defined, x must be more than or equal to zero.

Hope it's clear.

But what about taking negative root of a number?

So in this case, what if we take negative root of X and positive root of Y? In that case option A won't be sufficient...

1, square root X > square root Y Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD... 2, X^2>Y^2 here x= 1,y= 2 so 1 is not greater than 2, so

only A....

can we consider negative values in square root??

You cannot get sufficiency based only on one set of numbers. Thus theoretically testing only one set of numbers can give you an incorrect answer.

As for your other questions: the even roots from negative numbers are not defined for the GMAT, which means that for \(\sqrt{x}\) to be defined, x must be more than or equal to zero.

Hope it's clear.

But what about taking negative root of a number?

So in this case, what if we take negative root of X and positive root of Y? In that case option A won't be sufficient...

Where am I going wrong?

Thanks in advance! NB

I am assuming that you are talking about taking the negative square of a number? Something similar to , if given x^2 > y^2 ---> \(\sqrt{x^2} > \sqrt{y^2}\) ?

Note that for GMAT, square root of any positive number 'x' = \(\sqrt {x}\) \(\geq\) 0. There are no 'negative' square roots for GMAT.

• \(\sqrt{x^2}=|x|\), when x≤0, then \(\sqrt{x^2}=−x\) and when x≥0, then \(\sqrt{x^2}=x\)

• When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

So in this case, what if we take negative root of X and positive root of Y? In that case option A won't be sufficient...

Where am I going wrong?

Thanks in advance! NB[/quote]

I am assuming that you are talking about taking the negative square of a number? Something similar to , if given x^2 > y^2 ---> \(\sqrt{x^2} > \sqrt{y^2}\) ?

Note that for GMAT, square root of any positive number 'x' = \(\sqrt {x}\) \(\geq\) 0. There are no 'negative' square roots for GMAT.

• \(\sqrt{x^2}=|x|\), when x≤0, then \(\sqrt{x^2}=−x\) and when x≥0, then \(\sqrt{x^2}=x\)

• When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

Is x>y? (1) \(\sqrt{x}>\sqrt{y}\) The output of a square root is always a single positive root ; ONE UNIQUE POSITIVE ROOT VALUE \(\sqrt{X} >\sqrt{Y}\); therefore X>Y SUFFICIENT

(2) \(x^2>y^2\) For all practical purposes any number \(n^2\) can be seen as \(| n |\) and will always have a positive and negative root : TWO UNIQUE ROOT VALUES (+VE , -VE) Since X can be bigger or smaller than Y (depending upon the relative sign of x and y); Nothing can be said with confidence example \(-5^2 > 4^2\) but -5 < 4 example \(-3^2 < -4^2\) but -3> -4 INSUFFICIENT

ANSWER IS ONLY A
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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016.

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