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M02-19

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M02-19  [#permalink]

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New post 16 Sep 2014, 00:18
4
26
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (01:34) correct 65% (01:26) wrong based on 233 sessions

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Re M02-19  [#permalink]

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New post 16 Sep 2014, 00:18
4
6
Official Solution:


Is \(x \gt y^2\)?

(1) \(x \gt y+5\). Rewrite as \(x-y \gt 5\). 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) \(x^2-y^2=0\). Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) 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) \(x-y \gt 5 \ne 0\), then from (2) must be true that \(x+y=0\). So, \(x=-y\). Substitute \(x\) in (1): \(-y-y \gt 5\) 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: M02-19  [#permalink]

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New post 16 Sep 2014, 12:12
1
Bunuel wrote:
Official Solution:


(1) \(x \gt y+5\). Rewrite as \(x-y \gt 5\). 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) \(x^2-y^2=0\). Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) 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) \(x-y \gt 5 \ne 0\), then from (2) must be true that \(x+y=0\). So, \(x=-y\). Substitute \(x\) in (1): \(-y-y \gt 5\) 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: M02-19  [#permalink]

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New post 15 Nov 2014, 13:01
6
5
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: M02-19  [#permalink]

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New post 02 Dec 2014, 09:08
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: M02-19  [#permalink]

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New post 21 Jul 2016, 06:33
I think this is a high-quality question and I agree with explanation.
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Re: M02-19  [#permalink]

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New post 13 Jul 2017, 02:14
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: M02-19  [#permalink]

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New post 13 Jul 2017, 03:00
Prashant10692 wrote:
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.


(1)+(2) As from (1) \(x-y \gt 5 \ne 0\), then from (2) must be true that \(x+y=0\). So, \(x=-y\). Substitute \(x\) in (1): \(-y-y \gt 5\) from which we have that: \(y \lt -\frac{5}{2} \lt 0\), as \(x=-y\), then \(x \gt \frac{5}{2} \gt 0\).

Can you please tell me which part above is unclear? Thank you.
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Re: M02-19  [#permalink]

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New post 13 Sep 2017, 00:25
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: M02-19  [#permalink]

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New post 13 Sep 2017, 00:43
prathi123 wrote:
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.


From (2): either \(x-y=0\) or \(x+y=0\).
From (1): \(x-y \gt 5 \ne 0\). Notice that this rules out the first case from above.

Thus for (1)+(2): \(x+y=0\).

Hope it's clear.
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Re: M02-19  [#permalink]

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New post 24 Mar 2018, 03:35
Found this question extremely time consuming. Not because of the question's difficulty but for the varied number of cases need to be compared.
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Re: M02-19  [#permalink]

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New post 08 Apr 2018, 06:32
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.


Thanks for the alternative explanation. However can someone explain why only x = -y was plugged in? Is it the case because plugging in x=y will make statement 1 invalid/undefined?
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Re: M02-19  [#permalink]

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New post 07 Aug 2018, 03:20
is x>y^2?

1) x>y+5
x-y>5
x= 10 & y = 5, no.
x=6 & y =1 , yes.

option a alone not sufficient.

2) x^2 - y^2 = 0
(x-y)(x+y) = 0
x=y or x=-y not sufficient.

1 + 2
x+y=0 & x-y>5
solving these equations,
x> 2.5 & y <-2.5 & x=-y

which is sufficient. hence option c.

is my explanation correct?
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Re: M02-19 &nbs [#permalink] 07 Aug 2018, 03:20
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