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

It is currently 20 Nov 2018, 23:58

Close

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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

     November 22, 2018

     November 22, 2018

     10:00 PM PST

     11:00 PM PST

    Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
  • Free lesson on number properties

     November 23, 2018

     November 23, 2018

     10:00 PM PST

     11:00 PM PST

    Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 14 Jun 2017, 04:59
2
18
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

34% (01:45) correct 66% (02:06) wrong based on 275 sessions

HideShow timer Statistics

If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


SOLUTION (OE) IS HERE.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 23 Jun 2017, 11:33
1
6

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

General Discussion
Director
Director
User avatar
P
Joined: 18 Aug 2016
Posts: 625
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
GMAT ToolKit User Premium Member Reviews Badge
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 14 Jun 2017, 05:49
1
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis



So the question basically asks if Y=0

(x^2 - y^2)^(1/2) = x + y

Squaring both sides

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

(1) xy is not a square of an integer
Since \(-x \leq y \leq x\) and xy is not a square means x is not equal to y
But x+y > 0

(\(\sqrt{x^2 - y^2}\) is +ve) => x^2>y^2

If y = 0 then xy=0 which is a square of an integer hence y is not zero
and x cannot be 0 because y will be negative then as x>y and then x+y becomes negative but we know that x+y>0

Hence Sufficient

(2) Point (x,y) is not on x axis shows that y can be anything but 0. Sufficient

Hence both can solve
Answer D

Please correct me if i am wrong
_________________

We must try to achieve the best within us


Thanks
Luckisnoexcuse

PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 14 Jun 2017, 10:16
1
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 14 Jun 2017, 10:52
Intern
Intern
avatar
B
Joined: 12 Jun 2017
Posts: 9
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 15 Jun 2017, 06:42
niks18 wrote:
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E

It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 15 Jun 2017, 06:51
Mayur21 wrote:
niks18 wrote:
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


As per question stem -
\(\sqrt{x^2 - y^2} = x + y\). squaring it we get

\(x^2 - y^2= x^2 + y^2 + 2xy\) solving this we get
\(y^2+xy=0\) OR y(y+x) = 0
So the question asks us IS y=0 or y=-x

Statement 1: xy is not a square, implies that "y" is not equal to 0 but y can be equal to -x. For e.g x=-2, y=2, then xy = -4, which is not a perfect square as square cannot be negative, or any other value e.g x=2 y=3. Hence insufficient

Statement 2: this implies that x & y are not 0. but x and y can be any number, so y may or may not be equal to -x e.g (2,3) or (2,-2). Hence insufficient

Combining 1 & 2: we know that y is not equal to 0 but "y" and "x" can be any other number. Hence insufficient.

Option E

It is mentioned in the stem that -x<y<x. So y cannot be equal to -x.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app


It's less than or equal sign there: \(-x \leq y \leq x\), so y can equal to -x, as well as x. In fact -x=y=x=0 is also possible.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

SVP
SVP
User avatar
D
Joined: 26 Mar 2013
Posts: 1887
Reviews Badge CAT Tests
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 15 Jun 2017, 17:53
Bunuel wrote:
There are flaws in both solutions above, even in the one with correct answer.

P.S. Above question and similar question HERE are brand new questions by me and they seem to be quite tricky.



Hi Bunuel,

Those questions are awesome. You tagged both question as GMAT Club tests. Have you add them in the tests? i.e should I face them if I take mock exam in GMAT Club tests?
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7041
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 15 Jun 2017, 19:54
1
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


\(-x \leq y \leq x\) means x is positive or 0..
\(\sqrt{x^2 - y^2} = x + y\) ..
Square both sides..
\(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\)
So we have to find whether y=0 or x=-y...

I tells us that x is not equal to y..
But can be 0 or x=-y
Insuff
II tells us y is not equal to 0..
Insuff..

Combined
X can be = -y or not
Insufficient

E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Senior Manager
Senior Manager
User avatar
G
Joined: 19 Oct 2012
Posts: 327
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35
GMAT 2: 710 Q50 V38
GPA: 3.81
WE: Information Technology (Computer Software)
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 15 Jun 2017, 21:09
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis



\(\sqrt{x^2 - y^2} = x + y\) - > this result will come about when \(x^2 - y^2\) is either 0 or 1.
Let's jump on to the statement analysis.
1) xy is not a square of integer.

This means x and y are either different integers or same integers with different signs or they both are 0. Taking all the 3 scenarios,
When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true.
When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true.
When x=0 and y=0, \(\sqrt{x^2 - y^2} = x + y\) is true.

Hence insufficient.


2) Point (x, y) is NOT on x-axis

This implies y is not 0, but does not tell anything about x. Hence this statement in itself is insufficient.


Combining Stmt 1 and Stmt 2: Below two scenarios still hold :

When x=2 and y=-2, \(\sqrt{x^2 - y^2} = x + y\) is true.
When x=3 and y=2, \(\sqrt{x^2 - y^2} = x + y\) is not true.

Hence E.
_________________

Citius, Altius, Fortius

SVP
SVP
User avatar
D
Joined: 26 Mar 2013
Posts: 1887
Reviews Badge CAT Tests
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 16 Jun 2017, 01:46
chetan2u wrote:
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) xy is NOT a square of an integer
(2) Point (x, y) is NOT on x-axis


\(-x \leq y \leq x\) means x is positive or 0..
\(\sqrt{x^2 - y^2} = x + y\) ..
Square both sides..
\(x^2-y^2=(x+y)^2=x^2+y^2+2xy.........2y^2+2xy=0......2y(y+x) =0\)
So we have to find whether x=0 or x=-y...

I tells us that x is not equal to y..
But can be 0 or x=-y
Insuff
II tells us y is not equal to 0..
Insuff..

Combined
X can be = -y or not
Insufficient

E


I think you mean either y=o or x =-y

Also I have a question if y = 0, is not xy = 0 which is square of an integer zero??
Manager
Manager
avatar
B
Joined: 06 Nov 2016
Posts: 103
Location: India
GMAT 1: 710 Q50 V36
GPA: 2.8
Reviews Badge
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 21 Jun 2017, 00:35
Statement 1:
if y = 0, answer is yes
if y is not 0, answer is no

Statement 1 only tells us that x is not equal to y

Insuff

Statement 2:
tells us that y is not equal to 0
but ignores the possibility of x = y or x = -y

Insuff.

Combining - we know that neither x = y and nor y is 0
however x = -y is still not known

hence E
Manager
Manager
avatar
B
Joined: 06 Nov 2016
Posts: 103
Location: India
GMAT 1: 710 Q50 V36
GPA: 2.8
Reviews Badge
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 21 Jun 2017, 00:50
Bunuel - please provide detailed solution for the answer when OA is revealed,
This question is an absolute nightmare
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 23 Jun 2017, 11:34
SVP
SVP
User avatar
D
Joined: 26 Mar 2013
Posts: 1887
Reviews Badge CAT Tests
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 23 Jun 2017, 12:34
Bunuel wrote:

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.


Kudos to you Bunuel

it is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test?

I hope so

Thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 23 Jun 2017, 12:38
Mo2men wrote:
Bunuel wrote:

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.


Kudos to you Bunuel

it is really spot on. It is high quality question and answer. But do you plan to add in the GMATCLub tests to experience it in mock test?

I hope so

Thanks


Yes, all new questions will be added to the GMAT Club Tests.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
B
Joined: 26 Feb 2014
Posts: 7
GMAT 1: 700 Q49 V35
GPA: 4
Reviews Badge
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 21 Sep 2017, 00:24
Bunuel wrote:

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.


I get the answer as B. Can you tell me where I might be making a mistake here:
(b) Point (x, y) is NOT on x-axis
\(\sqrt{x^2 - y^2} = x + y\)
=> \(\sqrt{(x-y)(x+y)} = (x+y)\)
Dividing both sides by \(\sqrt{(x+y)}\), we get
\(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal.
Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50712
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 21 Sep 2017, 00:29
1
InfiniteLoop80 wrote:
Bunuel wrote:

OFFICIAL SOLUTION:



If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

First of all, \(-x \leq y \leq x\) ensures two things:
1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.
2. \(x+y\geq 0\), so the square root won't be equal to negative number.

Next, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by y, because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?


(1) xy is NOT a square of an integer.

If y = 0 were true, then xy would be 0, which is a square of an integer. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then xy would be -x^2, so negative, which cannot be a square of an integer. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

Thus, we have have two different answers to the question: Not sufficient.


(2) Point (x, y) is NOT on x-axis

If y = 0 were true, then point (x, y) would be ON the x-axis. So, we know that y ≠ 0.
However, if x = -y and x ≠ 0 were true, for example if x = -y = 1, then then point (x, y) would be (x, -x), so (positive, negative), which would mean that it's below x-axis. So, x = -y IS possible.
Of course, x can be not equal to -y, and the statemented still could hold true. For example, x = 2 and y =1.

(1)+(2) We could have the same cases: if x = -y and x ≠ 0, for example if x = -y = 1, then the answer is YES but if say x = 2 and y =1, then the answer is NO. Not sufficient.


Answer: E.


I get the answer as B. Can you tell me where I might be making a mistake here:
(b) Point (x, y) is NOT on x-axis
\(\sqrt{x^2 - y^2} = x + y\)
=> \(\sqrt{(x-y)(x+y)} = (x+y)\)
Dividing both sides by \(\sqrt{(x+y)}\), we get
\(\sqrt{(x-y)} = \sqrt{(x+y)}\)

If y = 0 , then x-y = x+y, otherwise they are not equal.
Since y != 0(since point (x,y) is not on x axis), x-y not equal to to x+y, hence \(\sqrt{(x-y)} != \sqrt{(x+y)}\) . So, went with B


Please re-read the solution, paying attention to the highlighted part.

As for your solution, you cannot divide by \(\sqrt{(x+y)}\) because it can be 0 for x = -y and we cannot divide by 0.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8869
Premium Member
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

Show Tags

New post 25 Sep 2018, 19:12
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y &nbs [#permalink] 25 Sep 2018, 19:12
Display posts from previous: Sort by

If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.