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If x and y are positive, is x^3 > y? [#permalink]
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Updated on: 02 Dec 2012, 03:58
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If x and y are positive, is x^3 > y? (1) \(\sqrt{x} > y\) (2) x > y
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Originally posted by udaymathapati on 30 Aug 2010, 09:38.
Last edited by Bunuel on 02 Dec 2012, 03:58, edited 2 times in total.
Edited the question.



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Re: Inequality DS [#permalink]
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30 Aug 2010, 10:31
udaymathapati wrote: If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y
My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS
Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3
Any other apporach. If x and y are positive, is x^3>y?NUMBER PLUGGING:(1) \(\sqrt{x}>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (2) \(x>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient. Answer: E. ALGEBRAIC APPROACH: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 \(y<x\)), so if we have this case answer is always YES: \(y<x^3\). But: For \(0<x<1\): \(0\)\(x^3\)\(x\)\(\sqrt{x}\)\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\) Answer: E. Hope it's clear.
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Re: Inequality DS [#permalink]
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31 Aug 2010, 04:07
When there are squaring or cubing is involved, always remember ve values as well as the values between 1 and 1



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Re: Inequality DS [#permalink]
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18 Mar 2011, 07:11
Bunuel wrote: udaymathapati wrote: If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y
My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS
Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3
Any other apporach. If x and y are positive, is x^3>y?NUMBER PLUGGING:(1) \(\sqrt{x}>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (2) \(x>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient. Answer: E. ALGEBRAIC APPROACH: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 \(y<x\)), so if we have this case answer is always YES: \(y<x^3\). But: For \(0<x<1\): \(0\)\(x^3\)\(x\)\(\sqrt{x}\)\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\) Answer: E. Hope it's clear. +1 for u Great Post Thanks Bunuel



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Re: Inequality DS [#permalink]
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18 Mar 2011, 21:25
x = 1/4, y = 1/8 From (1) 1/2 > 1/8, but 1/8 = 1/8 but (9)^3 > 2 and sqrt(9) > 2 In (2) also, the same is true. Answer  E.
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Re: If x and y are positive, is x^3 > y? [#permalink]
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Updated on: 30 May 2014, 15:37
udaymathapati wrote: If x and y are positive, is x^3 > y?
(1) \(\sqrt{x} > y\) (2) x > y My method without number plugging: 1) \(\sqrt{x}>y\) Square both sides. Since x & y are both positive, we know this doesn't change the sign at all (because \(f(x) = x^2\) is a strictly increasing function for \(x \geq 0\)). Thus \(x > y^2\). We're trying to get this in a form that resembles our original question, so we multiply each side by \(x^2\)(which we know is positive). Thus, \(x^3 > x^2y^2\), or that\(x^3 > y (x^2 y).\) Therefore, our question is TRUE if and only if \(x^2y\geq 1\). Since we don't know anything about the behavior of\(x^2y\), we mark 1) as INSUFFICIENT. 2) \(x>y.\) Multiplying both sides by \(x^2\), we get that \(x^3 > yx^2\). Thus, if \(x^2 \geq 1,\) we get TRUE, otherwise, we get FALSE. Since we have no information about \(x^2\), we mark this as INSUFFICIENT. Taking 1) and 2) together, we still have no information about \(x^2\) or \(x^2y\), so 1) & 2) are INSUFFICIENT. Answer: E.
Originally posted by eaze on 30 May 2014, 07:10.
Last edited by eaze on 30 May 2014, 15:37, edited 1 time in total.



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Re: Inequality DS [#permalink]
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30 May 2014, 14:46
Bunuel wrote: udaymathapati wrote: If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y
My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS
Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3
Any other apporach. If x and y are positive, is x^3>y?NUMBER PLUGGING:(1) \(\sqrt{x}>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (2) \(x>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient. Answer: E. ALGEBRAIC APPROACH: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 \(y<x\)), so if we have this case answer is always YES: \(y<x^3\). But: For \(0<x<1\): \(0\)\(x^3\)\(x\)\(\sqrt{x}\)\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\) Answer: E. Hope it's clear. What if the question says x and y are positive INTEGERS ??



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Re: Inequality DS [#permalink]
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30 May 2014, 15:40
ShajibAhmed wrote: What if the question says x and y are positive INTEGERS ??
If they were both integers, then the answer would be D: either of them are sufficient, since we'd know that\(x^2y \geq 1\) and \(x^2 \geq 1\)(see my solution above). The question doesn't state that, however, so we must consider all cases.



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Re: If x and y are positive, is x^3 > y? [#permalink]
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14 Feb 2017, 00:26
Bunuel wrote: udaymathapati wrote: If x and y are positive, is \(x^3\) > y? (1) \sqrt{x} > y (2) x > y
My approach: 1.\sqrt{x}>y ....>x>\(y^2\)....NS 2. x>y ....NS
Combine...\(y^2\)>x>y....> \(x^3\)>y. This can be verified by taking no like 4>3>2..where y=2 and x=3
Any other apporach. If x and y are positive, is x^3>y?NUMBER PLUGGING:(1) \(\sqrt{x}>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (2) \(x>y\) > if \(x=1\) and \(y=\frac{1}{2}\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{5}\) then the answer will be NO. Two different answers, hence not sufficient. (1)+(2) Both examples are valid for combined statements, so we still have two answers. Not sufficient. Answer: E. ALGEBRAIC APPROACH: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 \(y<x\)), so if we have this case answer is always YES: \(y<x^3\). But: For \(0<x<1\): \(0\)\(x^3\)\(x\)\(\sqrt{x}\)\(1\), so \(0<x^3<x<\sqrt{x}\). \(y\) is somewhere in green or red zone (as \(y<\sqrt{x}\) and \(y<x\)), so if we have this case answer is sometimes YES: \(y<x^3\) (if \(y\) is in green zone), and sometimes NO: \(x^3<y\) (if \(y\) is in red zone). In fact in this case \(y=x^3\) is also possible, for example when \(x=\frac{1}{2}\) and \(y=\frac{1}{8}\) Answer: E. Hope it's clear. You will get a better understanding by drawing the number line and plugging in the value.
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Re: If x and y are positive, is x^3 > y? [#permalink]
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