Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Nov 2012
Posts: 52

Is x > 0?
[#permalink]
Show Tags
Updated on: 01 Aug 2013, 04:43
Question Stats:
63% (01:29) correct 37% (01:38) wrong based on 296 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, 04:31.
Last edited by Zarrolou on 01 Aug 2013, 04:43, edited 2 times in total.
Edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 56251

Re: Is x > 0?
[#permalink]
Show Tags
06 May 2014, 08:52
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
_________________




Senior Manager
Joined: 24 Aug 2009
Posts: 457
Schools: Harvard, Columbia, Stern, Booth, LSB,

Re: Is x > 0?
[#permalink]
Show Tags
01 Aug 2013, 04:41
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, 07:04
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: 305

Re: Is x > 0?
[#permalink]
Show Tags
03 Aug 2013, 05:32
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
_________________



Manager
Joined: 04 Nov 2012
Posts: 52

Re: Is x > 0?
[#permalink]
Show Tags
04 Aug 2013, 03:53
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: 1647
Concentration: Finance

Re: Is x > 0?
[#permalink]
Show Tags
26 Apr 2014, 12:56
Shortcut LHS, RHS for this one anyone? Cheers! J



SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance

Re: Is x > 0?
[#permalink]
Show Tags
19 May 2014, 12:32
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: 56251

Re: Is x > 0?
[#permalink]
Show Tags
20 May 2014, 01:34
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?
_________________



NonHuman User
Joined: 09 Sep 2013
Posts: 11693

Re: Is x > 0?
[#permalink]
Show Tags
26 Sep 2018, 09:11
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.
_________________










