GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 26 May 2019, 17:13 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # M02-19

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 55277
M02-19  [#permalink]

### Show Tags

7
38 00:00

Difficulty:   95% (hard)

Question Stats: 37% (02:01) correct 63% (01:59) wrong based on 281 sessions

### HideShow timer Statistics

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

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

(2) $$x^2 - y^2 = 0$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 55277
Re M02-19  [#permalink]

### Show Tags

6
9
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
_________________
Manager  Joined: 18 Jan 2014
Posts: 164
Location: India
GMAT 1: 740 Q50 V40 Re: M02-19  [#permalink]

### Show Tags

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.
Manager  Joined: 04 Oct 2013
Posts: 165
Concentration: Finance, Leadership
GMAT 1: 590 Q40 V30 GMAT 2: 730 Q49 V40 WE: Project Management (Entertainment and Sports)
Re: M02-19  [#permalink]

### Show Tags

8
6
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.
_________________
learn the rules of the game, then play better than anyone else.
Manager  Joined: 10 Aug 2014
Posts: 58
GMAT 1: 600 Q44 V28 GMAT 2: 660 Q44 V35 Re: M02-19  [#permalink]

### Show Tags

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
Senior Manager  Joined: 31 Mar 2016
Posts: 376
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
WE: Operations (Commercial Banking)
Re: M02-19  [#permalink]

### Show Tags

I think this is a high-quality question and I agree with explanation.
Manager  G
Joined: 21 Mar 2017
Posts: 137
Location: India
GMAT 1: 560 Q48 V20 WE: Other (Computer Software)
Re: M02-19  [#permalink]

### Show Tags

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.
_________________
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
When nothing seem to help, I would go and look at a Stonecutter hammering away at his rock perhaps a hundred time without as much as a crack showing in it.
Yet at the hundred and first blow it would split in two.
And I knew it was not that blow that did it, But all that had gone Before
.
Math Expert V
Joined: 02 Sep 2009
Posts: 55277
Re: M02-19  [#permalink]

### Show Tags

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.
_________________
Intern  B
Joined: 13 Feb 2017
Posts: 35
Re: M02-19  [#permalink]

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 55277
Re: M02-19  [#permalink]

### Show Tags

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.
_________________
Manager  G
Joined: 25 Aug 2015
Posts: 51
Re: M02-19  [#permalink]

### Show Tags

1
Found this question extremely time consuming. Not because of the question's difficulty but for the varied number of cases need to be compared.
_________________
** When even your best effort fails, do you back down from chasing your dreams ??? **
Intern  B
Joined: 22 Jul 2015
Posts: 12
Location: Kuwait
Concentration: Entrepreneurship, Technology
GMAT 1: 640 Q42 V35 GMAT 2: 670 Q48 V34 GMAT 3: 710 Q47 V41 Re: M02-19  [#permalink]

### Show Tags

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?
Intern  B
Joined: 05 Jun 2017
Posts: 3
Re: M02-19  [#permalink]

### Show Tags

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?
Intern  B
Joined: 14 Oct 2018
Posts: 8
GMAT 1: 680 Q48 V34 Re M02-19  [#permalink]

### Show Tags

I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 24 Dec 2011
Posts: 44
Location: India
GPA: 4
WE: General Management (Health Care)
Re: M02-19  [#permalink]

### Show Tags

very decent question indeed...

is x>y^2?

1) x>y+5
by variable method. got multiple answers and ruled out.
x-y>5
x= 11 & y = 5, original question- No.
x=7 & y =1 , original question- Yes.

option a alone not sufficient.

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

got multiple answers..
x=y=2, the original question becomes No
x=0.5, y= -0.5, the original question becomes Yes

Combining both the statements
From st-1 x-y>5
which means in the st-2 the x-y=0 is not true. hence we take only
x+y=0
which means x=-y. substituting x=-y in the eq x-y>5,

-2y>5 => y< -5/2 inturn y is less than zero

x=-y means x is greater than zero

take any value of y which is less than -5/2 and x is always less than y^2
there's a definite answer and hence the answer is C Re: M02-19   [#permalink] 08 May 2019, 05:21
Display posts from previous: Sort by

# M02-19

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

Moderators: chetan2u, Bunuel Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.

#### MBA Resources  