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# M02 #19

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Manager
Joined: 26 May 2011
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Concentration: Entrepreneurship, Finance
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Re: M02 #19 [#permalink]

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14 Sep 2011, 17:21
I disagree with the OA.

x = +/- y.

The answer should be E IMO.

combining statement 1 & 2 we will have,

x > y+5, only possible when x is positive and y is negative. Because if both are positive and equal the inequality wouldn't hold.

Now we can chose some numbers, x= 2, y =-4
is x > y^2 ? NO

For x = 5, y = -1,

Is x>y^2 ? YES.

INSUFF. Therefore the answer should be E.

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Manager
Affiliations: University of Tehran
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Location: Iran (Islamic Republic of)
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Re: M02 #19 [#permalink]

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27 Sep 2011, 22:04
Bunuel wrote:
cnrnld wrote:
my opinion:

statement 2: x^2-y^2=0
we get 1) x=y, or x=-y; 2) x=y=0

then x>y^2 is false, statement 2 alone is sufficient.

choice B

Is $$x>y^2$$?

(1) $$x>y+5$$ --> $$x-y>5$$. 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) $$x^2-y^2=0$$ --> $$(x-y)(x+y)=0$$ --> 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) $$x-y>5\neq{0}$$, then from (2) must be true that $$x+y=0$$ --> so $$x=-y$$ --> substitute $$x$$ in (1) --> $$-y-y>5$$ --> $$y<-\frac{5}{2}<0$$, as $$x=-y$$, then $$x>\frac{5}{2}>0$$, so $$y^2$$ (or which is the same $$x^2$$) will always be more than $$x$$, thus the answer to the question "Is $$x>y^2$$" is NO. Sufficient.

To elaborate more as $$x=-y>0$$, the only chance for $$x>y^2$$ to hold true (or which is the same for $$x>x^2$$ to hold true) would be if $$x$$ is fraction ($$0<x<1$$). For example if $$x=\frac{1}{2}$$ and $$y=-\frac{1}{2}$$ then $$x=\frac{1}{2}>y^2=\frac{1}{4}$$. But the fact that $$x>\frac{5}{2}>0$$ rules out this option.

Hope it's clear.

+1, great explanation. 2.5 clarified!
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Senior Manager
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Re: M02 #19 [#permalink]

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27 Sep 2011, 22:21
Is $$x \gt y^2$$ ?

1. $$x \gt y + 5$$
2. $$x^2 - y^2 = 0$$

not sure that I am right. please let me know if my way of thinking is wrong/true

at first glance stmt 1 and 2 alone are not suff

now mix of 1 and 2

X>y+5 or x-y>5
x^2 - y^2 = 0 or (x-y)(x+y)=0

since x-y>5 x+y=0 or x=-y => so ,the answer to the question whether x>y^2 is NO
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Manager
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Re: M02 #19 [#permalink]

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28 Sep 2011, 00:24
I got E, still confused if C is the right answer

see for getting that y < 2.5 and x> 2.5, we have used both the eq.
I assume the following is the method by which that is derived::
x^2 = y^2------------- (a)
x > y + 5
squaring both sides
x^2 > y^2 + 10y + 25
using Eq (a) 25+ 10y < 0
y < 2.5

but there is an error in the above method, which is the squaring of the inequality equation.
squaring of inequality Eq can only be done if the signs of both sides are known.
Ex: read row wise in the table below: In all cases x > y + 5
x___y___ x^2___y^2
2 __-4___ 4 ___16 ---> x^2 < y^2
6 __ 0 ___36 ___ 0 ---> x^2 > y^2

so since the signs of x and y are unclear, I opted for E.

can anyone help me with this
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Re: M02 #19 [#permalink]

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30 Sep 2011, 00:05
Depaulian wrote:
I got E, still confused if C is the right answer

see for getting that y < 2.5 and x> 2.5, we have used both the eq.
I assume the following is the method by which that is derived::
x^2 = y^2------------- (a)
x > y + 5
squaring both sides
x^2 > y^2 + 10y + 25
using Eq (a) 25+ 10y < 0
y < 2.5

but there is an error in the above method, which is the squaring of the inequality equation.
squaring of inequality Eq can only be done if the signs of both sides are known.
Ex: read row wise in the table below: In all cases x > y + 5
x___y___ x^2___y^2
2 __-4___ 4 ___16 ---> x^2 < y^2
6 __ 0 ___36 ___ 0 ---> x^2 > y^2

so since the signs of x and y are unclear, I opted for E.

can anyone help me with this

Statement (1) alone is not sufficient. From x^2 = y^2 it transpires that |x| = |y| , and if |x| = |y| > 1 , then x cannot be greater than y^2; however, if x is equal to a positive fraction less than 1, then x necessarily is greater than y^2 (for example, 1/4 is greater than 1/16).

Statement (2) alone also is not sufficient. By plugging numerical values in x-5>y, if x is 7 and y is 1, then x is greater than y^2; on the other hand, if x is 10 and y is 4, then x is not greater than y^2. However, taking (1) and (2) together is sufficient; combining |x| = |y| from (1), and x-5>y from (2), yielding x-y>5, it necessarily follows that x is greater than 2.5 and y is less than -2.5, while the absolute values of x and y are equal. For example, if x equals 3, then y equals -3; if x equals 4, then y equals -4, and so forth. Thus x can never be greater than y^2, and the answer is a definite “no”; so, the correct answer is (C).

It is very interesting to note that a negative affirmative answer, i.e. one with a definite “no” to constitute a correct answer, started appearing in the GMAT only after Pearson Vue took over as administrators. Before 2006, when ETS were GMAC's administrators, the correct answer was an affirmative affirmative answer only - a definite “yes” was the only available correct choice.

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Re: M02 #19 [#permalink]

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04 Oct 2011, 05:37
jackychamp wrote:
Is $$x \gt y^2$$ ?

1. $$x \gt y + 5$$
2. $$x^2 - y^2 = 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

St1: When X=7 and Y=1, $$x \gt y^2$$, but when X=4 and Y= -3, $$x \lt y^2$$. not sufficient

St2: $$x^2 = y^2$$ When X=1 and Y=-1, then $$x = y^2$$, when X=-1 and Y= -1, $$x \lt y^2$$. in both cases $$x not \gt y^2$$. sufficient

B

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Manager
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Re: M02 #19 [#permalink]

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14 Nov 2011, 22:31
jackychamp wrote:
Is $$x \gt y^2$$ ?

1. $$x \gt y + 5$$
2. $$x^2 - y^2 = 0$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The key is testing cases.

1. 5,-1 does work vs. 2,-10 which doesn't - not sufficient
2. 2,-2 does work vs. 1,-1 doesn't work. Key is testing zeroes and one's.

Taken together we know that it can't be zero or one so that it must be 3,-3 as that meets test one and meets test 2. Making the statement above false.

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Re: M02 #19   [#permalink] 14 Nov 2011, 22:31

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# M02 #19

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