It is currently 15 Dec 2017, 21:35

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

Events & Promotions in June
Open Detailed Calendar

Is x>0? 1) x+1>0

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

Hide Tags

Expert Post
Math Revolution GMAT Instructor
User avatar
P
Joined: 16 Aug 2015
Posts: 4481

Kudos [?]: 3160 [0], given: 0

GPA: 3.82
Premium Member
Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 21 Nov 2017, 17:18
Expert's post
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:06) correct 35% (01:13) wrong based on 69 sessions

HideShow timer Statistics

[GMAT math practice question]
Is \(x>0\)?

1) \(x+1>0\)
2) \(|x+4| < |x-2|\)
[Reveal] Spoiler: OA

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo

Kudos [?]: 3160 [0], given: 0

Intern
Intern
avatar
B
Joined: 20 Feb 2017
Posts: 41

Kudos [?]: 4 [0], given: 435

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 07:19
From statement 1 :
x>-1
so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient
from statement 2
x<-1 (sufficient)
explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4)
when x>2
x+4-x+2>0 - no solution
when -4<x<2
x+4-(-(x-2))<0
=> x<-1
when x<-4
-x-4-(-(x-2))<0 again no solution
Hence B is the answer

Kudos [?]: 4 [0], given: 435

DS Forum Moderator
avatar
S
Joined: 21 Aug 2013
Posts: 570

Kudos [?]: 185 [0], given: 289

Location: India
Premium Member CAT Tests
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 07:45
1
This post was
BOOKMARKED
(1) First statement is obviously insufficient because x+1 > 0 tells us that x > -1. So x could lie between 0 and -1 or it could be >0.

(2) |x+4| > |x-2|
One way of looking at this statement is as follows:

|x-(-4)| < |x-2|

This tells us that 'Distance between x and -4' on the number line is lesser than 'Distance between x and 2'. Now which number lies exactly in the middle of -4 and 2 ? That number is -1 (3 steps away from -4 and 3 steps away from 2).
So, all numbers to the right of -1 will have the property that their distance from -4 will be greater than their distance from 2. So they wont satisfy the condition given in this statement.
But, all the numbers to the left of -1 will have the property that their distance from -4 will be less than their distance from 2. All these numbers are the only numbers which will satisfy the given inequality in this statement.

Thus on the basis of second statement alone we can conclude that x lies to the left of -1 on the number line. So x must be negative and hence cannot be >0. We get NO as the answer.

Hence B answer

Kudos [?]: 185 [0], given: 289

DS Forum Moderator
avatar
S
Joined: 21 Aug 2013
Posts: 570

Kudos [?]: 185 [0], given: 289

Location: India
Premium Member CAT Tests
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 07:52
Raksat wrote:
From statement 1 :
x>-1
so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient
from statement 2
x<-1 (sufficient)
explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4)
when x>2
x+4-x+2>0 - no solution
when -4<x<2
x+4-(-(x-2))<0
=> x<-1
when x<-4
-x-4-(-(x-2))<0 again no solution
Hence B is the answer


Good solution Raksat. May I point out though that in the last step we get -2 < 0, which is TRUE. This means all the values of x which are less than -4 will satisfy the given inequality. So x < -1, YES, and it also includes all values which are less than -4.

I think you already know this, I just thought writing 'no solution' probably doesnt fit in here. Hence pointing it out :)

Kudos [?]: 185 [0], given: 289

Intern
Intern
avatar
B
Joined: 20 Feb 2017
Posts: 41

Kudos [?]: 4 [0], given: 435

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 08:13
amanvermagmat wrote:
Raksat wrote:
From statement 1 :
x>-1
so x could be -0.5,-0.8,0 1,5 ,7 .... hence insufficient
from statement 2
x<-1 (sufficient)
explanation : we have 3 cases to check : (x>2),(-4<x<2) and (x<-4)
when x>2
x+4-x+2>0 - no solution
when -4<x<2
x+4-(-(x-2))<0
=> x<-1
when x<-4
-x-4-(-(x-2))<0 again no solution
Hence B is the answer

Good solution Raksat. May I point out though that in the last step we get -2 < 0, which is TRUE. This means all the values of x which are less than -4 will satisfy the given inequality. So x < -1, YES, and it also includes all values which are less than -4.

I think you already know this, I just thought writing 'no solution' probably doesnt fit in here. Hence pointing it out :)

yes... any value less than -4 will satisfy the given inequality and in turn answer the question stem.

Kudos [?]: 4 [0], given: 435

Intern
Intern
avatar
Joined: 22 Nov 2017
Posts: 1

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

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 09:12
Explanation for |x+4|<|x−2|:

Can there be any number to which we add 4 and it becomes less than the number that we get after subtracting 2 from it? NOT POSSIBLE.
This is only possible when x is negative and we compare absolute values. Thus, x has to be negative or x < 0.

Its good to solve the inequality and arrive at the conclusion methodically. But even if we don't do that, we can address the sufficiency question.

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

Expert Post
1 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5353

Kudos [?]: 6140 [1], given: 121

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 09:56
1
This post received
KUDOS
Expert's post
MathRevolution wrote:
[GMAT math practice question]
Is \(x>0\)?

1) \(x+1>0\)
2) \(|x+4| < |x-2|\)




Hi..
Just for those who would be looking for ranges in statement II..

1) x+1>0
x>-1...
So x could be 0.... Ans NO
X could be 3.... ANS Yes
Insuff

2) |x+4|<|x-2|
Square both sides..
\(x^2+8x+16<x^2-4x+4......12x<-12.....x<-1\)
Ans NO
Suff

B
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6140 [1], given: 121

Intern
Intern
avatar
B
Joined: 20 Feb 2017
Posts: 41

Kudos [?]: 4 [0], given: 435

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 10:14
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]
Is \(x>0\)?

1) \(x+1>0\)
2) \(|x+4| < |x-2|\)




Hi..
Just for those who would be looking for ranges in statement II..

1) x+1>0
x>-1...
So x could be 0.... Ans NO
X could be 3.... ANS Yes
Insuff

2) |x+4|<|x-2|
Square both sides..
\(x^2+8x+16<x^2-4x+4......12x<-12.....x<-1\)
Ans NO
Suff

B

i dont think we can square here .
consider x= 1 then 5^2>1^2
-3<x<5
squaring
0<x^2<25

Kudos [?]: 4 [0], given: 435

Expert Post
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5353

Kudos [?]: 6140 [0], given: 121

Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 10:56
Raksat wrote:
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]
Is \(x>0\)?

1) \(x+1>0\)
2) \(|x+4| < |x-2|\)




Hi..
Just for those who would be looking for ranges in statement II..

1) x+1>0
x>-1...
So x could be 0.... Ans NO
X could be 3.... ANS Yes
Insuff

2) |x+4|<|x-2|
Square both sides..
\(x^2+8x+16<x^2-4x+4......12x<-12.....x<-1\)
Ans NO
Suff

B

i dont think we can square here .
consider x= 1 then 5^2>1^2
-3<x<5
squaring
0<x^2<25[/quote]


Sorry I have not understood your query..
The equation is |x+4|<|x-2|

What the answer is x<-1
So x=-2 means |-2+4|<|-2-2|......2<4..
.ok
Test all values
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6140 [0], given: 121

Expert Post
Math Revolution GMAT Instructor
User avatar
P
Joined: 16 Aug 2015
Posts: 4481

Kudos [?]: 3160 [0], given: 0

GPA: 3.82
Premium Member
Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 18:41
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1)
\(x + 1 > 0\)
\(x > -1\)
Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2)
\(|x+4| < |x-2|
⟺|x+4|^2 < |x-2|^2
⟺(x+4)^2 < (x-2)^2
⟺x^2 + 8x + 16 < x^2 -4x + 4
⟺8x + 16 < -4x + 4
⟺12x < -12\)
\(⟺x < -1\)
The answer is ‘no’.
By CMT (Common Mistake Type) 1, ‘no’ is also an answer.
This is sufficient.

Therefore, the answer is B.


If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

Answer: B
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo


Last edited by MathRevolution on 23 Nov 2017, 11:54, edited 1 time in total.

Kudos [?]: 3160 [0], given: 0

Intern
Intern
avatar
B
Joined: 20 Feb 2017
Posts: 41

Kudos [?]: 4 [0], given: 435

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 20:59
chetan2u MathRevolution
My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?

Kudos [?]: 4 [0], given: 435

1 KUDOS received
DS Forum Moderator
avatar
S
Joined: 21 Aug 2013
Posts: 570

Kudos [?]: 185 [1], given: 289

Location: India
Premium Member CAT Tests
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 23:45
1
This post received
KUDOS
Raksat wrote:
chetan2u MathRevolution
My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?


Hi Raksat

I think we can square here because both sides are under modulus. Modulus of anything can never be negative. Lets say you are given that a non negative quantity a is less than a non negative quantity b. Since both are non-negative, we can safely say that square of a will be less than square of b. (you can take any two non negative quantities whether integers or fractions and check)
Now |x+4| is < |x-2|. Since both |x+4| and |x-2| are non negative (since they are absolute values) we can safely square both sides.

Kudos [?]: 185 [1], given: 289

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42618

Kudos [?]: 135771 [1], given: 12708

Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 22 Nov 2017, 23:50
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
Raksat wrote:
chetan2u MathRevolution
My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?


1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Check for more here: Manipulating Inequalities (adding, subtracting, squaring etc.).

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread


_________________

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

Kudos [?]: 135771 [1], given: 12708

1 KUDOS received
Intern
Intern
User avatar
B
Joined: 30 Sep 2017
Posts: 35

Kudos [?]: 12 [1], given: 26

Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
GPA: 3.7
WE: Engineering (Energy and Utilities)
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 25 Nov 2017, 18:40
1
This post received
KUDOS
I have a different approach to view Statement 2.
Please see and confirm my understanding is correct.

Bunuel wrote:
Raksat wrote:
chetan2u MathRevolution
My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?


1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Check for more here: Manipulating Inequalities (adding, subtracting, squaring etc.).

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread


Attachments

WhatsApp Image 2017-11-26 at 08.06.42.jpeg
WhatsApp Image 2017-11-26 at 08.06.42.jpeg [ 50.4 KiB | Viewed 467 times ]


_________________

If you like my post, motivate me by giving kudos...

Kudos [?]: 12 [1], given: 26

1 KUDOS received
Intern
Intern
User avatar
B
Joined: 30 Sep 2017
Posts: 35

Kudos [?]: 12 [1], given: 26

Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
GPA: 3.7
WE: Engineering (Energy and Utilities)
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 25 Nov 2017, 19:03
1
This post received
KUDOS
Dear Bunuel, MathRevolution
As per the Statement 1: x>-1 ahile Statement 2 ,x< -1. Doesnt it makes the question inconsistent.

Regards
MathRevolution wrote:
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1)
\(x + 1 > 0\)
\(x > -1\)
Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2)
\(|x+4| < |x-2|
⟺|x+4|^2 < |x-2|^2
⟺(x+4)^2 < (x-2)^2
⟺x^2 + 8x + 16 < x^2 -4x + 4
⟺8x + 16 < -4x + 4
⟺12x < -12\)
\(⟺x < -1\)
The answer is ‘no’.
By CMT (Common Mistake Type) 1, ‘no’ is also an answer.
This is sufficient.

Therefore, the answer is B.


If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

Answer: B

_________________

If you like my post, motivate me by giving kudos...

Kudos [?]: 12 [1], given: 26

Manager
Manager
User avatar
S
Status: Searching for something I've been searching..LOL
Joined: 14 Dec 2016
Posts: 64

Kudos [?]: 25 [0], given: 158

Location: India
Concentration: Healthcare, Operations
Schools: Ross '20
GMAT 1: 590 Q35 V42
GPA: 3.5
WE: Medicine and Health (Health Care)
GMAT ToolKit User CAT Tests
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 25 Nov 2017, 19:25
Barui wrote:
Dear Bunuel, MathRevolution
As per the Statement 1: x>-1 ahile Statement 2 ,x< -1. Doesnt it makes the question inconsistent.

Regards
MathRevolution wrote:
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer.

Condition 1)
\(x + 1 > 0\)
\(x > -1\)
Since the range of the question does not include the range of the condition, this is not sufficient.

Condition 2)
\(|x+4| < |x-2|
⟺|x+4|^2 < |x-2|^2
⟺(x+4)^2 < (x-2)^2
⟺x^2 + 8x + 16 < x^2 -4x + 4
⟺8x + 16 < -4x + 4
⟺12x < -12\)
\(⟺x < -1\)
The answer is ‘no’.
By CMT (Common Mistake Type) 1, ‘no’ is also an answer.
This is sufficient.

Therefore, the answer is B.


If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

Answer: B



I am no expert in quants but as far as I could understand.. The question stem is

*IS X<0* and we need a definite answer to this question.

Statement A does not give us a definite answer....
Whereas
Statement B gives us a definite answer....

That's what I could understand.

Correct me if I am wrong

Posted from my mobile device

Kudos [?]: 25 [0], given: 158

DS Forum Moderator
avatar
S
Joined: 21 Aug 2013
Posts: 570

Kudos [?]: 185 [0], given: 289

Location: India
Premium Member CAT Tests
Re: Is x>0? 1) x+1>0 [#permalink]

Show Tags

New post 25 Nov 2017, 22:04
Hi Barui

Your approach is correct.

Barui wrote:
I have a different approach to view Statement 2.
Please see and confirm my understanding is correct.

Bunuel wrote:
Raksat wrote:
chetan2u MathRevolution
My query is : Can we always square both the sides in an inequality involving modulus on both the sides as it is in this question. And why will the sign remain same after squaring ?


1. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we cannot square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

2. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Check for more here: Manipulating Inequalities (adding, subtracting, squaring etc.).

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread

Kudos [?]: 185 [0], given: 289

Re: Is x>0? 1) x+1>0   [#permalink] 25 Nov 2017, 22:04
Display posts from previous: Sort by

Is x>0? 1) x+1>0

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


GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.