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Is x > y? (1) x^2 > y (2) √x < y

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Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   Updated on: 19 Feb 2012, 18:39
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Is x > y?

(1) x^2 > y
(2) √x < y
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E

Originally posted by dvinoth86 on 19 Feb 2012, 08:00.
Last edited by dvinoth86 on 19 Feb 2012, 18:39, edited 1 time in total.
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   19 Feb 2012, 11:30
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Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.

(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that \(x\geq{0}\).

(1)+(2) \(\sqrt{x}<y<x^2\) --> both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\), but we can not say which one is greater. Not sufficient. For example: if \(x=y=4\) (\(\sqrt{4}<4<4^2\)) then the answer is NO but if \(x=4\) and \(y=3\) (\(\sqrt{4}<3<4^2\)) then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   20 Feb 2012, 00:18
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GMATPASSION wrote:
Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that \(x\geq{0}\).

(1)+(2) \(\sqrt{x}<y<x^2\) --> both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\), but we can not say which one is greater. Not sufficient. For example: if \(x=y=4\) (\(\sqrt{4}<4<4^2\)) then the answer is NO but if \(x=4\) and \(y=3\) (\(\sqrt{4}<3<4^2\)) then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.


Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.


First of all: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, number picking strategy can vary for different problems. Generally it's good to test negative/positive/zero as well as integer/fraction to get a YES and a NO answers. If you deal with two variables it's also helpful to test x<y and x>y in addition to the former.

As for this question: you don't really need to test the numbers for it, I just used them to demonstrate that the statements are not sufficient.

From (1)+(2): we have that \(\sqrt{x}<y<x^2\). Both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\) (\(x\) is between them because \(\sqrt{x}<x^2\), which means that \(x>1\)):
\(\sqrt{x}\)------\(x\)------\(x^2\), now \(y\) can be in the green range (answer YES) as well in the red range (answer NO). So, we can not say whether x>y.

Hope it's clear.
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   19 Feb 2012, 21:17
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Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that \(x\geq{0}\).

(1)+(2) \(\sqrt{x}<y<x^2\) --> both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\), but we can not say which one is greater. Not sufficient. For example: if \(x=y=4\) (\(\sqrt{4}<4<4^2\)) then the answer is NO but if \(x=4\) and \(y=3\) (\(\sqrt{4}<3<4^2\)) then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.


Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.
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Re: Is x > y? [#permalink] New post   18 Aug 2012, 22:43
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y


Show Spoiler
E


Can anyone explain a simple method to this could not follovv statement B

\

Bunuel if you can help please



1) Statement 1 only tells us that x is positive and nothing else. So insufficient

2) Statement 2 wants us to go through the process of squaring both sides to make the equation x<y^2, but we do not know anything about the sign so basically it would look like:

x^2 > y

and combined,

x<y^2 or x>y^2

x^2 > y
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Re: Is x > y? [#permalink] New post   Updated on: 18 Aug 2012, 23:04
[quote="venmic"]Is x > y?

(1) x^2 > y

(2) √x < y


Show Spoiler
E


Can anyone explain a simple method to this could not follovv statement B


(1) For \(x=-2 > y=-3, \, x^2=4>-3.\) But \(x=2 < y=3\), although \(x^2=4>y=3.\)
Not sufficient.

(2) From the given inequality it follows that \(x\) must be non-negative (because of the square root) and since \(y>\sqrt{x}\geq0\), necessarily \(y\) is positive.
Therefore, we can square the given inequality and get \(x<y^2.\)

For \(x=4, y=3, \,x=4<y^2=9,\) and \(x>y.\)
But if \(y=1,\) we cannot have simultaneously \(x<y^2=1\) and \(x>y=1.\)
Not sufficient.

(1) and (2) together:
Consider the two cases: \(x=2, y=3\) and \(x=4,y=3.\)
Again, not sufficient.

Answer E.

Originally posted by EvaJager on 18 Aug 2012, 22:56.
Last edited by EvaJager on 18 Aug 2012, 23:04, edited 1 time in total.
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Re: Is x > y? [#permalink] New post   18 Aug 2012, 23:03
hfbamafan wrote:
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y


Show Spoiler
E


Can anyone explain a simple method to this could not follovv statement B

\

Bunuel if you can help please



1) Statement 1 only tells us that x is positive and nothing else. So insufficient

2) Statement 2 wants us to go through the process of squaring both sides to make the equation x<y^2, but we do not know anything about the sign so basically it would look like:

x^2 > y

and combined,

x<y^2 or x>y^2

x^2 > y


(2) We know about the signs: \(x\) must be non-negative, otherwise the square root is not defined. Also, because the square root is non-negative, \(y\) must be positive. Therefore, in this case we can square the given inequality and obtain \(x<y^2.\)
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Re: Is x > y? [#permalink] New post   18 Aug 2012, 23:58
statement 2: in Gmat.... sqrt(x) means x is always positive .... therefore from statement 2 we know x is a positive number.

from statement 1 : x*x=y ..insufficient because we x and y can have any values....

combining 1 and 2:
sqrt(x) < y < x*x .... not possible to determine whether x> y
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   26 Jun 2021, 07:05
Bunuel wrote:
GMATPASSION wrote:
Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that \(x\geq{0}\).

(1)+(2) \(\sqrt{x}<y<x^2\) --> both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\), but we can not say which one is greater. Not sufficient. For example: if \(x=y=4\) (\(\sqrt{4}<4<4^2\)) then the answer is NO but if \(x=4\) and \(y=3\) (\(\sqrt{4}<3<4^2\)) then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.


Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.


First of all: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, number picking strategy can vary for different problems. Generally it's good to test negative/positive/zero as well as integer/fraction to get a YES and a NO answers. If you deal with two variables it's also helpful to test x<y and x>y in addition to the former.

As for this question: you don't really need to test the numbers for it, I just used them to demonstrate that the statements are not sufficient.

From (1)+(2): we have that \(\sqrt{x}<y<x^2\). Both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\) (\(x\) is between them because \(\sqrt{x}<x^2\), which means that \(x>1\)):
\(\sqrt{x}\)------\(x\)------\(x^2\), now \(y\) can be in the green range (answer YES) as well in the red range (answer NO). So, we can not say whether x>y.

Hope it's clear.


Hi Bunuel,

Pls can you tell us how we can conclude that x > 1 because \(\sqrt{x}\) < \(x^2\) please ?

Thanks
Krishnan
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   26 Jun 2021, 09:48
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KrishnanTN wrote:
Hi Bunuel,

Pls can you tell us how we can conclude that x > 1 because \(\sqrt{x}\) < \(x^2\) please ?

Thanks
Krishnan


\(\sqrt{x}<x^2\);

Square: \(x < x^4\):

Reduce by x: \(1 < x^3\). Here we know that x > 0 (because if x were negative \(\sqrt{x}\) would not be defined), so we can safely reduce by x.

Take the cube root: \(1 < x\).
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   26 Jun 2021, 10:53
Bunuel wrote:
KrishnanTN wrote:
Hi Bunuel,

Pls can you tell us how we can conclude that x > 1 because \(\sqrt{x}\) < \(x^2\) please ?

Thanks
Krishnan


\(\sqrt{x}<x^2\);

Square: \(x < x^4\):

Reduce by x: \(1 < x^3\). Here we know that x > 0 (because if x were negative \(\sqrt{x}\) would not be defined), so we can safely reduce by x.

Take the cube root: \(1 < x\).


I take it that the we are eliminating the case where x = 0, because \(\sqrt{x}\) cannot be lesser than x^2 if x were to be equal to 0.
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Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post   26 Jun 2021, 11:05
Bunuel wrote:
GMATPASSION wrote:
Bunuel wrote:
Is x > y?

(1) x^2 > y. Clearly insufficient: if x=2 and y=3 then the answer is NO but if x=2 and y=1 then the answer is YES.
(2) √x < y. Also insufficient: if x=4 and y=5 then the answer is NO but if x=4 and y=3 then the answer is YES. Notice that since x is under the square root sign then it must be true that \(x\geq{0}\).

(1)+(2) \(\sqrt{x}<y<x^2\) --> both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\), but we can not say which one is greater. Not sufficient. For example: if \(x=y=4\) (\(\sqrt{4}<4<4^2\)) then the answer is NO but if \(x=4\) and \(y=3\) (\(\sqrt{4}<3<4^2\)) then the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.


Its not always possible to take examples like you have shown for each statement. I mean sometimes the variable values just doesn't fit. Solving it by use of abstract maths is tough.

Ho do we tackle this situation. Do we have a strategy on how to pick numbers faster for testing.


First of all: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, number picking strategy can vary for different problems. Generally it's good to test negative/positive/zero as well as integer/fraction to get a YES and a NO answers. If you deal with two variables it's also helpful to test x<y and x>y in addition to the former.

As for this question: you don't really need to test the numbers for it, I just used them to demonstrate that the statements are not sufficient.

From (1)+(2): we have that \(\sqrt{x}<y<x^2\). Both \(x\) and \(y\) are between \(\sqrt{x}\) and \(x^2\) (\(x\) is between them because \(\sqrt{x}<x^2\), which means that \(x>1\)):
\(\sqrt{x}\)------\(x\)------\(x^2\), now \(y\) can be in the green range (answer YES) as well in the red range (answer NO). So, we can not say whether x>y.

Hope it's clear.


Hi Bunuel,

Is the following solution correct please ?
Is x >y ? [Let us plot x = y. This is a straight line passing through the origin at 45Degrees to the x axis.]

1) x^2 > y - This is a parabola, symmetrical around the y axis.
juxtapose the parabola and the straight line. Not sufficient.

2) x^.5 ,y - A half parabola. Juxtapose x=y and this. Not sufficient.

1 + 2 --> THese 3 curves still cannot answer the question.
Is this approach okay please ?
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Re: Is x > y? (1) x^2 > y (2) x < y [#permalink] New post   28 Jun 2023, 11:20
Can this be a logically correct approach?

Statement 1 and Statement 2: Insufficient

Together:

Subtract the equations:
X^2 - Y^2 > Y - X
(X + Y) (X - Y) > -(-Y+X)
=>X+Y>-1. Cannot create X-Y>0 from this combination therefore insufficient.
Ans E
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Re: Is x > y? (1) x^2 > y (2) x < y [#permalink] New post   20 Sep 2023, 10:32
I cam across the same question but in statement B the option is ( root of x is > y )
and the answer mentioned is option c :- both statement together sufficient

Bunuel can you please explain this question with the above mentioned context
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Re: Is x > y? (1) x^2 > y (2) x < y [#permalink] New post   20 Sep 2023, 10:35
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OmkarSheth008 wrote:
I cam across the same question but in statement B the option is ( root of x is > y )
and the answer mentioned is option c :- both statement together sufficient

Bunuel can you please explain this question with the above mentioned context


That question is discussed here: https://gmatclub.com/forum/is-x-y-1-x-2 ... 80724.html
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Re: Is x > y? (1) x^2 > y (2) x < y [#permalink]
20 Sep 2023, 10:35
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