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# Is x > y?

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24 Jul 2003, 05:23
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Is x>y?

(1) $$\sqrt{x}>\sqrt{y}$$
(2) $$x^2>y^2$$ -
[Reveal] Spoiler: OA

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KL

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25 Jul 2003, 00:09
Lynov Konstantin wrote:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2

All we know from (1) is that X and Y are positive (i.e. it could be an integer or a fraction). Consider an example:

X=4, Y=9 => sqrt(X)<sqrt(Y);
X=1/4, Y=1/9 => sqrt(X)>sqrt(Y);

Therefore, (1) alone is clearly not sufficient.

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.
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25 Jul 2003, 01:07
Lynov Konstantin wrote:
Lynov Konstantin wrote:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2

All we know from (1) is that X and Y are positive (i.e. it could be an integer or a fraction). Consider an example:

X=4, Y=9 => sqrt(X)<sqrt(Y);
X=1/4, Y=1/9 => sqrt(X)>sqrt(Y);

Therefore, (1) alone is clearly not sufficient.

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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25 Jul 2003, 01:19
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You should start from data sets to answer the question, not vice versa.

(1) square root X > square root Y

X and Y are nonnegative

sqrt 9> sqrt 4 => 3>2 => X>Y
sqrt 1/4 > sqrt 1/9 => 1/2>1/3 => X>Y
sqrt 100 > sqrt 1/9 => 10>1/3 => X>Y
sqrt 1/4 > sqrt 0 => 1/2>0 => X>Y

More than enough

(2) clearly not sufficient because of even powers

9>4 => 3>2 => X>Y OR
9>4 => -3<2 => X<Y

Finally, A.
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Re: An old one: Is X>Y ? (1) square root X > square root Y [#permalink]

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14 Jan 2012, 22:28
Explaination by stolyar is perfect.

Sqrt of X is greater than X if X < 1 but here the comparison is between 2 different numbers X and Y. So A is sufficient.
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Re: Is x>y? (1) square root x>square root y (2) x^2>y^2 [#permalink]

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15 Jan 2012, 04:08
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Konstantin Lynov wrote:
An old one:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2

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.

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.

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Re: Is x > y? [#permalink]

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11 Oct 2013, 02:37
pls correct my approach in solving this problem..

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??
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Re: Is x > y? [#permalink]

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11 Oct 2013, 05:36
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kanusha wrote:
pls correct my approach in solving this problem..

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.
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Re: Is x > y? [#permalink]

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07 Jan 2015, 05:28
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Re: Is x > y? [#permalink]

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26 Sep 2015, 12:55
Bunuel wrote:
kanusha wrote:
pls correct my approach in solving this problem..

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?

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Re: Is x > y? [#permalink]

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26 Sep 2015, 13:35
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nehabhasin wrote:
Bunuel wrote:
kanusha wrote:
pls correct my approach in solving this problem..

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?

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.

2 points to note (copied from math-number-theory-88376.html):

• $$\sqrt{x^2}=|x|$$, 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.

Hope this helps.
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Re: Is x > y? [#permalink]

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02 Oct 2015, 10:26
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?

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.

2 points to note (copied from math-number-theory-88376.html):

• $$\sqrt{x^2}=|x|$$, 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.

Hope this helps.[/quote]

Got it! Thanks!
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Re: Is x > y? [#permalink]

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18 Jul 2016, 06:18
Konstantin Lynov wrote:
Is x>y?

(1) $$\sqrt{x}>\sqrt{y}$$
(2) $$x^2>y^2$$ -

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

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Re: Is x > y?   [#permalink] 18 Jul 2016, 06:18
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