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Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
 08 Sep 2010, 01:38
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I thought this was a really tough question ! Is x > y ? (1) \sqrt{x} > y(2) x^3 > y
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shrouded1 wrote: I thought this was a really tough question !
Is x > y ?
(1) \sqrt{x} > y
(2) x^3 > y Is x>y?(1) \sqrt{x}>y --> if x=4 and y=1 then the answer will be YES but if x=\frac{1}{4} and y=\frac{1}{3} then the answer will be NO. Two different answers, hence not sufficient. Note that from this statement we can derive that x\geq{0} because an expression under the square root cannot be negative. (2) x^3>y --> if x=4 and y=1 then the answer will be YES but if x=2 and y=3 then the answer will be NO. Two different answers, hence not sufficient. (1)+(2) From (1) we have that x\geq{0}. Now, \sqrt{x}, x, x^3 can be positioned on a number line only in 2 ways: 1. For 1\leq{x}: ------\sqrt{x}----x----x^3, so 1\leq{\sqrt{x}}\leq{x}\leq{x^3} (the case \sqrt{x}=x=x^3 is when x=1). y is somewhere in green zone (as y<\sqrt{x} and y<x^3), so if we have this case answer is always YES: y<x. 2. For 0\leq{x}<1: 0----x^3----x----\sqrt{x}----1, so 0\leq{x^3}\leq{x}\leq{\sqrt{x}}. y is somewhere in green zone (as y<\sqrt{x} and y<x^3), so if we have this case answer is always YES: y<x. So in both cases y<x. Sufficient. Answer: C. Hope it's clear.
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Bunuel.... you seem to love number theory! I really am amazed at your patience. Good job. You already have enough Kudos. 
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interesting question.......forgot to consider fractions first.......now its clear.....
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That solution is so simple yet I couldn't figure it out. I knew that each statement alone was insufficient, but I couldn't figure out whether both together were sufficient. I kept trying to plug in numbers and it just never worked. Never did it occur to me to put both scenarios on a number line and just use the statements to prove it. I swear the math on the GMAT really makes you think in different ways and the solutions are so easy we make it harder than it needs to be.
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Nice explanation ....
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shrouded1 wrote: I thought this was a really tough question !
Is x > y ?
(1) \sqrt{x} > y
(2) x^3 > y （1）x^2>y^4 (2) x^3 > y if both 1and 2 are true, then x>0, if y<=0, then x>y; if y>0, we make x,y have the same power by (1)*（2） x^5>y^5, so x>y; so x>y.
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Great question Thanks for the post and for the explaination
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Again there is a wonderfully simple explanation for this one using graphs. x>y, sqrt(x)>y, x^3>y all three represents the region below the graph for all three cases. We need to answer is x>y  (1) sqrt(x)>y You are below the yellow line does not imply you are below the purple line. Insufficient (2) x^3>y Now x need not be just positive, but looking at the graph is enough to conclude this is not sufficient. Being below the blue line does not imply being below the purple line (1+2) Now x>0 since we are using sqrt(x) You are below the blue line and the yellow line both To satisfy both, you must always be below the purple line Answer is (C)
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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.
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Agreed It's just that the way I was taught algebra, there was a lot of focus on graphs. I find it way more intuitive than algebraic manipulation, more straight forward than plugging values and in almost all cases faster especially for very simple functions like these Posted from my mobile device 
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shrouded1 wrote: Agreed It's just that the way I was taught algebra, there was a lot of focus on graphs. I find it way more intuitive than algebraic manipulation, more straight forward than plugging values and in almost all cases faster especially for very simple functions like these Posted from my mobile device Even I was taught in same way. But post JEE, I never used them. I will work one day on graphs for sure. It is the best approach.
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Re: DS question : need help [#permalink]
28 Oct 2010, 14:30
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Hi Bunuel!, do you have similar questions? They would be very helpful to be sure that we have learned this  Thanks!
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shrouded1 wrote: I thought this was a really tough question !
Is x > y ?
(1) \sqrt{x} > y
(2) x^3 > y One of those gorgeous questions that seem so simple at first but surprise you later... Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!) Statement (1): If I can say that x >= \sqrt{x} for all values of x, then I can say that x > y. The green line shows me the region where x >= \sqrt{x} but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient. I also understand from this statement that x >= 0. Statement (2): If I can say that x >= x^3 for all values of x, then I can say that x > y. The green lines show me the region where x >= x^3 but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient. Attachment: 
Ques.jpg [ 9.04 KiB | Viewed 9138 times ]
Using both together, I know x >= 0. If 0 <= x <= 1, then we know x >= x^3. Since statement (2) says that x^3 > y, I can say that x > y. If x > 1, then we know x > \sqrt{x}. Since statement (1) says that \sqrt{x} > y, I can deduce that x > y. For all possible values of x, we can say x > y. Sufficient. Answer (C).
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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.
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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. ofcourse, it helps a lot bunnel.....thank you,,,my science background is killing me........
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satishreddy wrote:
hey karishma,,,,we can also rewrite sq root X>M as X>M2 , by squaring both sides to make things simple,,,,,,,,,
How about providing the solution with your suggested modifications? Is always great to see different takes on the same question!
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Kudos to shrouded1 and Bunuel. Excellent Post and super excellent solution.
Thanks both of you.
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Re: DS in inequality [#permalink]
12 Apr 2012, 04:56
Hi, To Prove : X> Y Statement 1 : sqrt (x) > y From this statement X is always positive hence X> Y 4 > 2 ,16> 4, etc Statement 2 : cube root (x) > y X can be negative & smaller than y eg -8 < -4 Thus A alone is sufficient. Hope this helps though I'm not 100% sure if I'm correct
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Re: DS in inequality
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