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Re M0219 [#permalink]
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15 Sep 2014, 23:18
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Official Solution:Is \(x \gt y^2\)? (1) \(x \gt y+5\). Rewrite as \(xy \gt 5\). Clearly insufficient, for example: if \(x=1\) and \(y=10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient. (2) \(x^2y^2=0\). Rewrite as \((xy)(x+y)=0\): so either \(xy=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient. (1)+(2) As from (1) \(xy \gt 5 \ne 0\), then from (2) must be true that \(x+y=0\). So, \(x=y\). Substitute \(x\) in (1): \(yy \gt 5\) from which we have that: \(y \lt \frac{5}{2} \lt 0\), as \(x=y\), then \(x \gt \frac{5}{2} \gt 0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x \gt y^2\)" is NO. Sufficient. To elaborate more as \(x=y \gt 0\), the only chance for \(x \gt y^2\) to hold true (or which is the same for \(x \gt x^2\) to hold true) would be if \(x\) is fraction (\(0 \lt x \lt 1\)). For example if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) then \(x=\frac{1}{2} \gt y^2=\frac{1}{4}\). But the fact that \(x \gt \frac{5}{2} \gt 0\) rules out this option. Answer: C
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Re: M0219 [#permalink]
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16 Sep 2014, 11:12
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Bunuel wrote: Official Solution:
(1) \(x \gt y+5\). Rewrite as \(xy \gt 5\). Clearly insufficient, for example: if \(x=1\) and \(y=10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient. (2) \(x^2y^2=0\). Rewrite as \((xy)(x+y)=0\): so either \(xy=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient. (1)+(2) As from (1) \(xy \gt 5 \ne 0\), then from (2) must be true that \(x+y=0\). So, \(x=y\). Substitute \(x\) in (1): \(yy \gt 5\) from which we have that: \(y \lt \frac{5}{2} \lt 0\), as \(x=y\), then \(x \gt \frac{5}{2} \gt 0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x \gt y^2\)" is NO. Sufficient. To elaborate more as \(x=y \gt 0\), the only chance for \(x \gt y^2\) to hold true (or which is the same for \(x \gt x^2\) to hold true) would be if \(x\) is fraction (\(0 \lt x \lt 1\)). For example if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) then \(x=\frac{1}{2} \gt y^2=\frac{1}{4}\). But the fact that \(x \gt \frac{5}{2} \gt 0\) rules out this option.
Answer: C Just a thought that last line makes more sense with, >> "But the fact that x \gt \frac{5}{2} \gt 1 rules out this option. Thanks for the explanation, really useful.



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Re: M0219 [#permalink]
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15 Nov 2014, 12:01
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Is \(x > y^2\)? 1) x > y+5 Instead of working with the inequality, assume that x = y+5 and keep in mind that x can have a slightly greater magnitude than that estimated (for instance +0,1) or a definitely greater one (for instance +5). Pick some values for y and compute x. If y = 1; x = 6 (precisely anything > 6) then x is greater than \(y^2\). Pick y = 6; x = 1 then \(y^2\) is greater than x. Not sufficient. 2) \(x^2  y^2 = 0\) → x = ± y. Pick some values for x. If x = 5; y = ± 5 then x is smaller than \(y^2\). If x = 0,5; y = ± 0,5 then x is greater than \(y^2\). Not sufficient. 1+2) plug x=y into the first inequality, you will get that y < 2,5. Say that y = 2,6 then x =2,6 and x is always smaller than \(y^2\). If two numbers are equal the question translates into: is \(x > x^2\)? Which happens only when 0 < x < 1. Sufficient → answer C.
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Re: M0219 [#permalink]
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02 Dec 2014, 08:08
gmat6nplus1 wrote: Is \(x > y^2\)?
1) x > y+5 Instead of working with the inequality, assume that x = y+5 and keep in mind that x can have a slightly greater magnitude than that estimated (for instance +0,1) or a definitely greater one (for instance +5). Pick some values for y and compute x. If y = 1; x = 6 (precisely anything > 6) then x is greater than \(y^2\). Pick y = 6; x = 1 then \(y^2\) is greater than x. Not sufficient.
2) \(x^2  y^2 = 0\) → x = ± y. Pick some values for x. If x = 5; y = ± 5 then x is smaller than \(y^2\). If x = 0,5; y = ± 0,5 then x is greater than \(y^2\). Not sufficient.
1+2) plug x=y into the first inequality, you will get that y < 2,5. Say that y = 2,6 then x =2,6 and x is always smaller than \(y^2\). If two numbers are equal the question translates into: is \(x > x^2\)? Which happens only when 0 < x < 1.
Sufficient → answer C. The combining 1+2 part well explained.. Thanks



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Re: M0219 [#permalink]
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21 Jul 2016, 05:33
I think this is a highquality question and I agree with explanation.



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Re: M0219 [#permalink]
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13 Jul 2017, 01:14
Hi Bunuel, Could you please explain from where the term y<5/2<0 came. I cant understand from where the term less than 0 came from.
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Re: M0219 [#permalink]
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13 Jul 2017, 02:00



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Re: M0219 [#permalink]
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12 Sep 2017, 23:25
Hi Bunuel,
Can you please explain how did you arrive at [b] As from (1) x−y>5≠0, 'then from (2) must be]true that x+y=0'. I am not able to get this part. Thanks in advance.



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Re: M0219 [#permalink]
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12 Sep 2017, 23:43










