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Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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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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Re: Is x > y ? [#permalink]
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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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Re: Is x > y ? [#permalink]
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08 Sep 2010, 22:50
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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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Re: Is x > y ? [#permalink]
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interesting question.......forgot to consider fractions first.......now its clear.....
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Re: Is x > y ? [#permalink]
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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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Re: Is x > y ? [#permalink]
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16 Sep 2010, 21:24
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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Re: Is x > y ? [#permalink]
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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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Re: Is x > y ? [#permalink]
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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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Re: Is x > y ? [#permalink]
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09 Oct 2010, 10:28
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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Re: Is x > y ? [#permalink]
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09 Oct 2010, 11:15
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]
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Re: Is x > y ? [#permalink]
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28 Oct 2010, 14:28
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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Re: Is x > y ? [#permalink]
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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 49540 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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Re: Is x > y ? [#permalink]
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29 Oct 2010, 14:47
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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 nonnegative (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 nonnegative. 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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Re: Is x > y ? [#permalink]
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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 nonnegative (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 nonnegative. 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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Re: Is x > y ? [#permalink]
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30 Oct 2010, 04:52
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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Re: DS in inequality [#permalink]
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12 Apr 2012, 03:56
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: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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13 Apr 2012, 10:12
Hi Karishma, Quote: 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).
from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap! Please help me with this1 Thanks



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Re: Is x > y ? (1) x^(1/2)>y (2) x^3>y [#permalink]
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18 Apr 2012, 02:28
shankar245 wrote: Hi Karishma, Quote: 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).
from the diagrams you have put up , how can we derive that X>=0 cos from both the diagrams the regions do not over lap! Please help me with this1 Thanks We know that x >= 0 because statement 1 tells us that \(\sqrt{x} > y\). \(\sqrt{x}\) can be defined only if x is non negative. Check out this post for a detailed explanation of this solution: http://www.veritasprep.com/blog/2011/08 ... question/
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Re: Is x > y ? [#permalink]
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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 nonnegative (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 nonnegative. 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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