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Is x > y ? (1) x^(1/2)>y (2) x^3>y

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Re: Is x > y ? [#permalink] New post 09 Oct 2012, 04:27
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,


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.

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.

Hope it helps.


Hi few observations please correct me if Im wrong ->

\sqrt{x} > y -> cannot square this but I can always cube both sides

y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??
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Re: Is x > y ? [#permalink] New post 09 Oct 2012, 05:37
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Jp27 wrote:
Bunuel wrote:
satishreddy wrote:
hey bunnel.....can we also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,


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.

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.

Hope it helps.


Hi few observations please correct me if Im wrong ->

\sqrt{x} > y -> cannot square this but I can always cube both sides

y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ??



\sqrt{x} > y -> cannot square this but I can always cube both sides - YES (although it won't help much, the square root will stay)
y > \sqrt{x} - Can i square this? Rational behind this is right hand side of the inequality is +ve so left must be to, hence this can be written as y^2>x ?? YES, but the right hand side always non-negative (it can be 0)
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 05 Nov 2012, 09:00
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 05 Nov 2012, 20:58
Expert's post
Maurice wrote:
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E


You might want to re-consider what happens when you take both statements together. Check out the solutions on page 1 or check this link:
http://www.veritasprep.com/blog/2011/08 ... -question/
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 06 Nov 2012, 04:49
Expert's post
Maurice wrote:
Is x > y ?

a). x2 > y
Gives 2 conditions x > y & x < y
eg:- Consider x=2 and y=1, then x2 > y and x > y
Consider x= -4 and y = 1, then x2 > y and x < y
Insufficient.

b) sqrt x < y
ie., x < y2, again as above gives 2 conditions x > y & x < y
Insufficient.

C) Form a we have x > y & x < y, and form b we have x > y & x < y, literally both are same
Insufficient.
Ans . E


Answer to the question is C, not E. Check the solutions on page 1 and 2. For example: is-x-y-1-x-1-2-y-2-x-3-y-100636.html#p777179

Hope it helps.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 16 Jan 2013, 04:36
It totally helped me to setup a chart to test the values out:

y < x
(a) increasing integer
(b) increasing fraction

y > x
(c) decreasing fraction
(d) decreasing integer

Test this out on statement (1) and statement (2)...
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 22 Jan 2013, 00:35
shrouded1 wrote:
I thought this was a really tough question !

Is x > y ?

(1) \sqrt{x} > y
(2) x^3 > y


I notice usually these types test the fractions... if x and y are swapping signs...

1. \sqrt{x} > y==>x > y^2

Let x>y: 2 > 1 (It works!)

Let x<y: 1/3 > (1/4)^2 (It works!)
Let x<y: -1/3 ? 1/16 (It doesn't work)

INSUFFICIENT!

2. x^3 > y

Let x>y: (2)^3 > 1 (It works!)

Let x<y: 1/27 ? 1/4 (It doesn't work!)
Let x<y: -1/27 > -1/4 (It works)

INSUFFICIENT!

Combine: eq in Statement 1 and 2 works for x>y

Answer: C
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Re: DS Inequality [#permalink] New post 05 Jun 2013, 07:48
Expert's post
Manhnip wrote:
Is x > y?

(1) √ x > y

(2) x^3 > y

Please provide an easy method to solve this problem in less than 2 mins


The number properties change when the numbers move beyond the range of 0 and 1. Draw a number line and try to check the values of \sqrt{x} and x^3.
Both the statements explain only about one range, but as soon as one combines both the statements, the range becomes quite clear and hence C becomes the answer.
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Re: Is x > y ? [#permalink] New post 27 Dec 2013, 08:42
gurpreetsingh wrote:
Statement 1: \sqrt{x} > y ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient.

Statement 2: x^3 > y take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient.

take statement 1 and 2 together. Now the answer could be either C or E.

Either y is -ve or postive. If y is -ve then x >y always holds true.

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y.

Hence C.

The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains.


Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

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Re: Is x > y ? [#permalink] New post 01 Jan 2014, 22:24
Expert's post
jlgdr wrote:
gurpreetsingh wrote:
Statement 1: \sqrt{x} > y ....take x = 4, y = 1 (yes); take x = 1/4 and y = 1/3 (no) hence not sufficient.

Statement 2: x^3 > y take x = 2 and y = 3 (no) , take x = 2, y = 1 (yes) hence not sufficient.

take statement 1 and 2 together. Now the answer could be either C or E.

Either y is -ve or postive. If y is -ve then x >y always holds true.

If y is +ve then x>y^2 and x^3>y => x^6 > y^2

divide both the inequalities: x^5 >1 => x >1. Since square root of x (> 1) is greater than y => x>y.

Hence C.

The graphical approach is good, but its essential to understand that how the inequalities behaves in different domains.


Yeah kinda tough indeed

Hey gupreet or anyone how did you get to this? If y is +ve then x>y^2 and x^3>y => x^6 > y^2

Like from both statements to the x^6 > y^2?

Thanks

Cheers!
J :)

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If y is positive and x^3>y, it means x^3 is also positive (since it is greater than y which is positive).
If both sides of an inequality are positive, you can square the inequality.

x^3>y
(x^3)^2>y^2
x^6 > y^2
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Is x > y? (1): sqrt(x) > y, (2): x^3 > y [#permalink] New post 19 Mar 2014, 14:25
Is x > y?

(1) sqrt(x) > y
(2) x^3 > y

Hi, I want to know if we can solving this without picking numbers, please.
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Re: Is x > y? (1): sqrt(x) > y, (2): x^3 > y [#permalink] New post 20 Mar 2014, 00:37
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink] New post 18 Apr 2014, 03:04
shrouded1 wrote:
I thought this was a really tough question !

Is x > y ?

(1) \sqrt{x} > y
(2) x^3 > y


Statement I is insufficient:

x = 4 and y = 3 then x is greater than y
x = 1/4 and y = 1/3 then x is not greater than y

Statement II is insufficient:
x = 4 and y = 2 then x is greater than y
x = -1/4 and y = -1/2 then x is not greater than y

Combining both we can clearly see that:
x^1/2 > y
x^3 > y

If we plug in positive fractions or positive numbers each time x is greater than y. Hence answer is C.
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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y   [#permalink] 18 Apr 2014, 03:04
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