M02 #19 : Retired Discussions [Locked] - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 16:13

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M02 #19

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 26 May 2011
Posts: 160
Concentration: Entrepreneurship, Finance
GPA: 3.22
Followers: 3

Kudos [?]: 26 [0], given: 10

Re: M02 #19 [#permalink]

### Show Tags

14 Sep 2011, 16:21
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.
Manager
Affiliations: University of Tehran
Joined: 06 Feb 2011
Posts: 203
Location: Iran (Islamic Republic of)
Grad GPA: 4
Concentration: Marketing
Schools: Wharton
GMAT 1: 680 Q45 V38
GPA: 4
WE: Marketing (Retail)
Followers: 8

Kudos [?]: 37 [0], given: 57

Re: M02 #19 [#permalink]

### Show Tags

27 Sep 2011, 21:04
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.

Answer: C.

Hope it's clear.

+1, great explanation. 2.5 clarified!
_________________

Ambition, Motivation and Determination: the three "tion"s that lead to PERFECTION.

World! Respect Iran and Iranians as they respect you! Leave the governments with their own.

Senior Manager
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 21

Kudos [?]: 322 [0], given: 73

Re: M02 #19 [#permalink]

### Show Tags

27 Sep 2011, 21:21
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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Manager
Status: Dreaming High
Joined: 24 Apr 2011
Posts: 61
Location: India
Concentration: Operations, Strategy
GMAT 1: 720 Q50 V36
GPA: 3.28
WE: Project Management (Manufacturing)
Followers: 0

Kudos [?]: 13 [0], given: 14

Re: M02 #19 [#permalink]

### Show Tags

27 Sep 2011, 23:24
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
_________________

[caption=]Remember: Anything that can go wrong, will go wrong.[/caption]

Intern
Joined: 25 Oct 2009
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: M02 #19 [#permalink]

### Show Tags

29 Sep 2011, 23:05
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.
Intern
Joined: 24 May 2010
Posts: 46
Followers: 0

Kudos [?]: 5 [0], given: 6

Re: M02 #19 [#permalink]

### Show Tags

04 Oct 2011, 04:37
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
Manager
Joined: 16 Sep 2010
Posts: 223
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Followers: 9

Kudos [?]: 112 [0], given: 2

Re: M02 #19 [#permalink]

### Show Tags

14 Nov 2011, 21:31
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.
Re: M02 #19   [#permalink] 14 Nov 2011, 21:31

Go to page   Previous    1   2   [ 27 posts ]

Similar topics Replies Last post
Similar
Topics:
m02 q 19 1 17 Dec 2011, 19:26
6 m02 #20 26 17 Jan 2009, 21:17
15 m02 #21 28 14 Nov 2008, 12:33
40 m02#24 14 05 Nov 2008, 07:21
22 m02#8 16 03 Nov 2008, 15:24
Display posts from previous: Sort by

# M02 #19

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.