December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Nov 2012
Posts: 52

Is x > 0?
[#permalink]
Show Tags
Updated on: 01 Aug 2013, 03:43
Question Stats:
64% (01:08) correct 36% (01:17) wrong based on 281 sessions
HideShow timer Statistics
Is \(x > 0\)? (1) \(x + 3 = 4x  3\) (2) \(x + 1 = 2x  1\) Now, I approached the question in the following manner:
Statement 1: x+3>=0 always So,
4x3>=0 so X>=3/4
Sufficient
Statement 2
similarly 2X1>=0 x>1/2 Sufficient
So: D each statement is enough.
Can you tell me whether I am missing something conceptually? Or a case in which this approach might fail?
Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??
Thanks..
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by 12bhang on 01 Aug 2013, 03:31.
Last edited by Zarrolou on 01 Aug 2013, 03:43, edited 2 times in total.
Edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 51218

Re: Is x > 0?
[#permalink]
Show Tags
06 May 2014, 07:52
jlgdr wrote: Shortcut LHS, RHS for this one anyone? Cheers! J Is \(x > 0\)? (1) \(x + 3 = 4x  3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. (2) \(x + 1 = 2x  1\). The smae here: LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(2x1\geq{0}\) > \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient. Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.) Similar question to practice: isx01x34x32x32x127978.html
_________________
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? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Joined: 24 Aug 2009
Posts: 469
Schools: Harvard, Columbia, Stern, Booth, LSB,

Re: Is x > 0?
[#permalink]
Show Tags
01 Aug 2013, 03:41
12bhang wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x + 1 = 2x – 1 Statement 1If x + 3 >= 0 > x+3 = 4x3 >3x=6>x=2 If x + 3 < 0 > x3 = 4x3 >5x=0>x=0 x=0 is not possible because we assumed that x+3<0 or x<3. Thus the only possible value of x is 2. Sufficient Statement 2If x + 1 >= 0 > x+1 = 2x1 >x=2>x=2 If x + 1 < 0 > x1 = 2x1 >3x=0>x=0 x=0 is not possible because we assumed that x+1<0 or x<1. Thus the only possible value of x is 2. Sufficient Answer D
_________________
If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth Game Theory
If you have any question regarding my post, kindly pm me or else I won't be able to reply



Manager
Joined: 04 Nov 2012
Posts: 52

Re: Is x > 0?
[#permalink]
Show Tags
01 Aug 2013, 06:04
fameatop wrote: 12bhang wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x + 1 = 2x – 1 Statement 1If x + 3 >= 0 > x+3 = 4x3 >3x=6>x=2 If x + 3 < 0 > x3 = 4x3 >5x=0>x=0 x=0 is not possible because we assumed that x+3<0 or x<3. Thus the only possible value of x is 2. Sufficient Statement 2If x + 1 >= 0 > x+1 = 2x1 >x=2>x=2 If x + 1 < 0 > x1 = 2x1 >3x=0>x=0 x=0 is not possible because we assumed that x+1<0 or x<1. Thus the only possible value of x is 2. Sufficient Answer D Ok. that is the correct conventional way. Here's what i thought. x+3 will always be either > or = to 0. So , 4x3>=0 So, x>3/4, which is greater than 0. So sufficient. statement 2: x+1>=0 So , 2x1>=0 or x>=1/2 which is greater than 0 so sufficient D What I want to know is whether there is any loophole in my approach. I.e am i missing out on some hidden value of the modulus or something and whether it is better to solve for X and then check whether it falls in the range. Please help. Thanks



Senior Manager
Joined: 10 Jul 2013
Posts: 315

Re: Is x > 0?
[#permalink]
Show Tags
03 Aug 2013, 04:32
12bhang wrote: Is \(x > 0\)? (1) \(x + 3 = 4x  3\) (2) \(x + 1 = 2x  1\) Now, I approached the question in the following manner:
Statement 1: x+3>=0 always So,
4x3>=0 so X>=3/4
Sufficient
Statement 2
similarly 2X1>=0 x>1/2 Sufficient
So: D each statement is enough.
Can you tell me whether I am missing something conceptually? Or a case in which this approach might fail?
Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??
Thanks.. this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions....... some good process already members publishes here, use them......... thanks
_________________
Asif vai.....



Manager
Joined: 04 Nov 2012
Posts: 52

Re: Is x > 0?
[#permalink]
Show Tags
04 Aug 2013, 02:53
Asifpirlo wrote: 12bhang wrote: Is \(x > 0\)? (1) \(x + 3 = 4x  3\) (2) \(x + 1 = 2x  1\) Now, I approached the question in the following manner:
Statement 1: x+3>=0 always So,
4x3>=0 so X>=3/4
Sufficient
Statement 2
similarly 2X1>=0 x>1/2 Sufficient
So: D each statement is enough.
Can you tell me whether I am missing something conceptually? Or a case in which this approach might fail?
Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??
Thanks.. this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions....... some good process already members publishes here, use them......... thanks Hi, Thanks for your insight. I agree that the equation does not hold true at the given values. However, the question does not need us to find the actual values. merely proving that the value of x>0 is enough. This will be proved, as the LHS in both statements are in the modulus and will return values that are greater than or equal to zero.



SVP
Joined: 06 Sep 2013
Posts: 1720
Concentration: Finance

Re: Is x > 0?
[#permalink]
Show Tags
26 Apr 2014, 11:56
Shortcut LHS, RHS for this one anyone? Cheers! J



SVP
Joined: 06 Sep 2013
Posts: 1720
Concentration: Finance

Re: Is x > 0?
[#permalink]
Show Tags
19 May 2014, 11:32
Bunuel wrote: jlgdr wrote: Shortcut LHS, RHS for this one anyone? Cheers! J Is \(x > 0\)? (1) \(x + 3 = 4x  3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. (2) \(x + 1 = 2x  1\). The smae here: LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(2x1\geq{0}\) > \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient. Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.) Similar question to practice: isx01x34x32x32x127978.htmlIs it because you have 2x  1 and therefore 2x>=1, x>=1/2? What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios Please advice Cheers! J



Math Expert
Joined: 02 Sep 2009
Posts: 51218

Re: Is x > 0?
[#permalink]
Show Tags
20 May 2014, 00:34
jlgdr wrote: Bunuel wrote: jlgdr wrote: Shortcut LHS, RHS for this one anyone? Cheers! J Is \(x > 0\)? (1) \(x + 3 = 4x  3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. (2) \(x + 1 = 2x  1\). The smae here: LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(2x1\geq{0}\) > \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient. Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.) Similar question to practice: isx01x34x32x32x127978.htmlIs it because you have 2x  1 and therefore 2x>=1, x>=1/2? What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios Please advice Cheers! J We could still solve \(x + 1 = 2x + 1\) with this approach to get x. LHS is an absolute value, which is always nonnegative, so RHS must also be nonnegative > \(2x+1\geq{0}\) > \(x\geq{\frac{1}{2}}\). For, \(x\geq{\frac{1}{2}}\), \(x+1>0\), hence \(x + 1 = x + 1\) > \(x + 1 = 2x + 1\) > \(x=0\). But if it were \(x  1 = 2x + 1\), then getting \(x\geq{\frac{1}{2}}\) from LHS, wouldn't be useful to determine whether \(x  1 = x  1\) or \(x  1 = (x  1)\) because \(x1\) could be positive as well as negative for different x'es which are more than or equal to 1/2. Therefore we should use usual method of solving for \(x  1 = 2x + 1\), not RHS/LHS. So, this method is not universal and can be applied only when nonmodulus part defines x in such way that helps to get the sign of absolute value expression on the other side. Does this make sense?
_________________
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? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 9164

Re: Is x > 0?
[#permalink]
Show Tags
26 Sep 2018, 08:11
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




Re: Is x > 0? &nbs
[#permalink]
26 Sep 2018, 08:11






