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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]

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19 Feb 2012, 08:00
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Is x > y?

(1) x^2 > y
(2) √x < y
[Reveal] Spoiler: OA

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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]

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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.

Hope it's clear.
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink]

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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.

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]

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19 Feb 2012, 21:17
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.

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]

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18 Aug 2012, 22:43
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y

[Reveal] 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]

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18 Aug 2012, 22:56
[quote="venmic"]Is x > y?

(1) x^2 > y

(2) √x < y

[Reveal] 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.

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Last edited by EvaJager on 18 Aug 2012, 23:04, edited 1 time in total.

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Re: Is x > y? [#permalink]

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18 Aug 2012, 23:03
hfbamafan wrote:
venmic wrote:
Is x > y?

(1) x^2 > y

(2) √x < y

[Reveal] 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]

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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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21 Jan 2013, 03:14
Is x > y?

(1) x^2 > y
(2) √x < y
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Re: Is x > y? [#permalink]

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21 Jan 2013, 03:22
monir6000 wrote:
Is x > y?

(1) x^2 > y
(2) √x < y

Merging similar topics. Please refer to the solutions above.

P.S. PLEASE PROVIDE OA'S FOR THE QUESTIONS!
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink]

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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink]

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