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M02-19

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:18
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Difficulty:

95% (hard)

Question Stats:

33% (01:39) correct 67% (01:25) wrong based on 215 sessions

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Is $$x \gt y^2$$?

(1) $$x \gt y + 5$$

(2) $$x^2 - y^2 = 0$$
[Reveal] Spoiler: OA

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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 $$x-y \gt 5$$. 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.

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16 Sep 2014, 11:12
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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.

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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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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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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21 Jul 2016, 05:33
I think this is a high-quality question and I agree with explanation.
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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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13 Jul 2017, 02:00
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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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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12 Sep 2017, 23:43
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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