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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 19 Feb 2012, 07:00
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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, 17:39, 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, 10:30
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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 19 Feb 2012, 20:17
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.

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? (1) x^2 > y (2) √x < y [#permalink] New post 19 Feb 2012, 23:18
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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? [#permalink] New post 18 Aug 2012, 21:43
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] New post 18 Aug 2012, 21:56
[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.

Answer E.
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Last edited by EvaJager on 18 Aug 2012, 22:04, edited 1 time in total.
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Re: Is x > y? [#permalink] New post 18 Aug 2012, 22:03
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] New post 18 Aug 2012, 22:58
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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Is x > y? [#permalink] New post 21 Jan 2013, 02:14
Is x > y?

(1) x^2 > y
(2) √x < y
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Re: Is x > y? [#permalink] New post 21 Jan 2013, 02:22
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Re: Is x > y? (1) x^2 > y (2) √x < y [#permalink] New post 17 Jul 2014, 11:01
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Re: Is x > y? (1) x^2 > y (2) √x < y   [#permalink] 17 Jul 2014, 11:01
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