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Updated on: 25 Jan 2022, 05:38
Originally posted by udaymathapati on 30 Aug 2010, 09:38.
Last edited by Bunuel on 25 Jan 2022, 05:38, edited 3 times in total.
Last edited by Bunuel on 25 Jan 2022, 05:38, edited 3 times in total.
Edited the question.
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E
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If x and y are positive, is x^3 > y?
(1) \(\sqrt{x} > y\)
(2) x > y
(1) \(\sqrt{x} > y\)
(2) x > y
30 Aug 2010, 10:31
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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.
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.
udaymathapati
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.
