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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 ??

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) _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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

Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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

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.
_________________

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

Kudos rain!

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.