It is currently 21 Mar 2018, 05:35

### 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

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

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

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5062
GMAT 1: 800 Q59 V59
GPA: 3.82

### Show Tags

21 Nov 2017, 18:18
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

68% (01:05) correct 32% (01:13) wrong based on 74 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.
"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 20 Feb 2017 Posts: 55 Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 08:19 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 DS Forum Moderator Joined: 22 Aug 2013 Posts: 893 Location: India Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 08:45 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 DS Forum Moderator Joined: 22 Aug 2013 Posts: 893 Location: India Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 08:52 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 Manager Joined: 20 Feb 2017 Posts: 55 Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 09:13 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. Intern Joined: 22 Nov 2017 Posts: 1 Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 10:12 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. Math Expert Joined: 02 Aug 2009 Posts: 5724 Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 10:56 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 GMAT online Tutor Manager Joined: 20 Feb 2017 Posts: 55 Re: Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 11:14 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 Math Expert Joined: 02 Aug 2009 Posts: 5724 Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 11:56 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 GMAT online Tutor Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5062 GMAT 1: 800 Q59 V59 GPA: 3.82 Is x>0? 1) x+1>0 [#permalink] ### Show Tags 22 Nov 2017, 19:41 1 This post received KUDOS Expert's post => 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. "Only$79 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Last edited by MathRevolution on 23 Nov 2017, 12:54, edited 1 time in total.
Manager
Joined: 20 Feb 2017
Posts: 55
Re: Is x>0? 1) x+1>0 [#permalink]

### Show Tags

22 Nov 2017, 21:59
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 ?
DS Forum Moderator
Joined: 22 Aug 2013
Posts: 893
Location: India
Re: Is x>0? 1) x+1>0 [#permalink]

### Show Tags

23 Nov 2017, 00:45
1
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.
Math Expert
Joined: 02 Sep 2009
Posts: 44383
Re: Is x>0? 1) x+1>0 [#permalink]

### Show Tags

23 Nov 2017, 00:50
1
KUDOS
Expert's post
2
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

_________________
Intern
Joined: 30 Sep 2017
Posts: 36
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

25 Nov 2017, 19:40
2
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 [ 50.4 KiB | Viewed 610 times ]

_________________

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

Intern
Joined: 30 Sep 2017
Posts: 36
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

25 Nov 2017, 20:03
1
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)
$$|x+4| < |x-2| ⟺|x+4|^2 < |x-2|^2 ⟺(x+4)^2 < (x-2)^2 ⟺x^2 + 8x + 16 < x^2 -4x + 4 ⟺8x + 16 < -4x + 4 ⟺12x < -12$$
$$⟺x < -1$$
By CMT (Common Mistake Type) 1, ‘no’ is also an answer.
This is sufficient.

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.

_________________

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

Manager
Status: Searching for something I've been searching..LOL
Joined: 14 Dec 2016
Posts: 64
Location: India
Concentration: Healthcare, Operations
Schools: Ross '20
GMAT 1: 590 Q35 V42
GPA: 3.5
WE: Medicine and Health (Health Care)
Re: Is x>0? 1) x+1>0 [#permalink]

### Show Tags

25 Nov 2017, 20:25
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)
$$|x+4| < |x-2| ⟺|x+4|^2 < |x-2|^2 ⟺(x+4)^2 < (x-2)^2 ⟺x^2 + 8x + 16 < x^2 -4x + 4 ⟺8x + 16 < -4x + 4 ⟺12x < -12$$
$$⟺x < -1$$
By CMT (Common Mistake Type) 1, ‘no’ is also an answer.
This is sufficient.

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.

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
DS Forum Moderator
Joined: 22 Aug 2013
Posts: 893
Location: India
Re: Is x>0? 1) x+1>0 [#permalink]

### Show Tags

25 Nov 2017, 23:04
Hi Barui

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

Re: Is x>0? 1) x+1>0   [#permalink] 25 Nov 2017, 23:04
Display posts from previous: Sort by