Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Oct 2011
Posts: 117
Location: India

Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
22 Feb 2012, 19:59
Question Stats:
43% (01:24) correct 57% (01:20) wrong based on 1589 sessions
HideShow timer Statistics
Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Encourage me by pressing the KUDOS if you find my post to be helpful. Help me win "The One Thing You Wish You Knew  GMAT Club Contest"http://gmatclub.com/forum/theonethingyouwishyouknewgmatclubcontest140358.html#p1130989




Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: [tricky one] Is x > 0? [#permalink]
Show Tags
22 Feb 2012, 22:02




Intern
Joined: 29 Aug 2011
Posts: 20

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
25 Feb 2012, 10:45
Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
a. x+3 = 4x 3 x+3 = x + 3 for x + 3 >= 0 > x > 3. Solving for x, x + 3 = 4x 3 > 6 = 3x > x = 2 ( solution accepted since x > 3)
 x + 3 = (x +3 ) for x + 3 < 0 > x < 3. Solving for x, x  3 = 4x 3 5x = 0 > x = 0 ( solution discarded as x is not < 3)
A is sufficient.
b. x 3 = 2x 3 x 3 = x 3 for x3 >= 0 > x >= 3 x 3 = (x 3) for x3 < 0 > x < 3
2x 3  = 2x 3 for 2x 3 >= 0 > x >=3/2 2x 3  = (2x 3) for 2x 3 < 0 > x < 3/2
So we have 3 ranges
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)
Let me know if I have solved the question correctly. I know the process is lengthy.
One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
Please discuss.
Regards




Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
25 Feb 2012, 10:52
saxenaashi wrote: Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
a. x+3 = 4x 3 x+3 = x + 3 for x + 3 >= 0 > x > 3. Solving for x, x + 3 = 4x 3 > 6 = 3x > x = 2 ( solution accepted since x > 3)
 x + 3 = (x +3 ) for x + 3 < 0 > x < 3. Solving for x, x  3 = 4x 3 5x = 0 > x = 0 ( solution discarded as x is not < 3)
A is sufficient.
b. x 3 = 2x 3 x 3 = x 3 for x3 >= 0 > x >= 3 x 3 = (x 3) for x3 < 0 > x < 3
2x 3  = 2x 3 for 2x 3 >= 0 > x >=3/2 2x 3  = (2x 3) for 2x 3 < 0 > x < 3/2
So we have 3 ranges
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)
Let me know if I have solved the question correctly. I know the process is lengthy.
One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
Please discuss.
Regards It seems that you understand this method very well. Everything is correct except the red parts: (x3) = (2x 3) > x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient. Hope it's clear.
_________________
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



Intern
Joined: 29 Aug 2011
Posts: 20

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
25 Feb 2012, 11:07
Bunuel wrote: saxenaashi wrote: Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
a. x+3 = 4x 3 x+3 = x + 3 for x + 3 >= 0 > x > 3. Solving for x, x + 3 = 4x 3 > 6 = 3x > x = 2 ( solution accepted since x > 3)
 x + 3 = (x +3 ) for x + 3 < 0 > x < 3. Solving for x, x  3 = 4x 3 5x = 0 > x = 0 ( solution discarded as x is not < 3)
A is sufficient.
b. x 3 = 2x 3 x 3 = x 3 for x3 >= 0 > x >= 3 x 3 = (x 3) for x3 < 0 > x < 3
2x 3  = 2x 3 for 2x 3 >= 0 > x >=3/2 2x 3  = (2x 3) for 2x 3 < 0 > x < 3/2
So we have 3 ranges
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)
Let me know if I have solved the question correctly. I know the process is lengthy.
One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
Please discuss.
Regards It seems that you understand this method very well. Everything is correct except the red parts: (x3) = (2x 3) > x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient. Hope it's clear. Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation. Regards



Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
25 Feb 2012, 11:15
saxenaashi wrote: Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation. Regards A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (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 can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. 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 \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). So in statement (2) since both parts of expression are nonnegative we can safely apply squaring. Hope it helps.
_________________
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



Intern
Joined: 26 Jan 2012
Posts: 1

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
27 Feb 2012, 23:09
Well, I have a little bit another solution for this problem Consider (1): if x>3 then x+3=4x3, so x=2, sufficient if x<3, then x3=4x3, which Leeds to answer x=0, which is wrong because we consider only x<3 Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficient In your logic, you gave a mistake, because you've missed the root. Posted from GMAT ToolKit



Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
27 Feb 2012, 23:30
terance wrote: Well, I have a little bit another solution for this problem Consider (1): if x>3 then x+3=4x3, so x=2, sufficient if x<3, then x3=4x3, which Leeds to answer x=0, which is wrong because we consider only x<3 Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficient In your logic, you gave a mistake, because you've missed the root. Posted from GMAT ToolKitWelcome to GMAT Club. Unfortunately your answer is not correct: OA for this question is A, not D (you can see it under the spoiler in the initial post). (2) x – 3 = 2x – 3 has two roots x=0 and x=2 (just substitute these values to see that they both satisfy the given equation), so you can not get the single numerical value of x, which makes this statement insufficient. You can refer to above solutions for two different approaches of how to get these roots for (2). Please ask if anything remains unclear.
_________________
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



Manager
Joined: 27 Oct 2011
Posts: 158
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
WE: Account Management (Consumer Products)

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
15 Apr 2012, 15:38
A simple solution for this is solve for x in each statement, but remember when you solve for x with abs value you must make the term positive and negative. in both cases you will find x = +# and zero. plug the numbers back into the equation and you will find zero will not fit the equation so x must be > 0
_________________
DETERMINED TO BREAK 700!!!



Manager
Joined: 30 Jun 2011
Posts: 235

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
17 Apr 2012, 06:19
dvinoth86 wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos Remember: When you have  on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have  on only one side you MUST VERIFY the answer by putting them back in eqn(1) x + 3 = 4x – 3 a) x + 3 = 4x – 3 => x=2 .. VALID b) (x + 3) = 4x – 3 => x=0 .. INVALID Hence Sufficient (2) x – 3 = 2x – 3 => x = 0 or 6 .. INVALID Hence InSufficient



Intern
Joined: 09 Apr 2012
Posts: 2

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
17 Apr 2012, 08:19
Is this a correct way to approach 2) x – 3 = 2x – 3 ? Step 1: (x3) = (2x3) => x = 0 Substitute x = 0 in x – 3 = 2x – 3 => VALID
Step 2: (x3) = (2x3) => x = 2 Substitute x = 2 in x – 3 = 2x – 3 => VALID
So 2) x – 3 = 2x – 3 is NOT SUFFICIENT



Manager
Joined: 30 Jun 2011
Posts: 235

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
17 Apr 2012, 08:41
kirncha wrote: Is this a correct way to approach 2) x – 3 = 2x – 3 ? Step 1: (x3) = (2x3) => x = 0 Substitute x = 0 in x – 3 = 2x – 3 => VALID
Step 2: (x3) = (2x3) => x = 2 Substitute x = 2 in x – 3 = 2x – 3 => VALID
So 2) x – 3 = 2x – 3 is NOT SUFFICIENT Yes but you can skip the substitution part, read my statement above



Intern
Joined: 02 Apr 2012
Posts: 4

Re: [tricky one] Is x > 0? [#permalink]
Show Tags
18 Apr 2012, 07:46
Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > 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. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. Pretty slick answer. Saves a lot of time.



Manager
Joined: 04 Dec 2011
Posts: 68

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
09 May 2013, 13:38
Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device
_________________
Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!
1 Kudos = 1 thanks Nikhil



Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
10 May 2013, 02:01
nikhil007 wrote: Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device If you substitute x=0 in x + 3 = 4x – 3 you'll get: LHS=x + 3=3 and RHS=4x – 3=3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation. When expanding x+3: When x<3, then x+3=(x+3), so in this case we'll have (x+3)=4x3 > x=0 > discard this value since 0 is not less than 3 (we consider the range when x<3). When x>=3, then x+3=x+3, so in this case we'll have x+3=4x3 > x=2 > this value of x is OK since 2>3. So, x + 3 = 4x  3 has only one root, x=2. Hope it's clear.
_________________
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



Manager
Joined: 04 Dec 2011
Posts: 68

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
11 May 2013, 06:07
Bunuel wrote: nikhil007 wrote: Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device If you substitute x=0 in x + 3 = 4x – 3 you'll get: LHS=x + 3=3 and RHS=4x – 3=3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation. When expanding x+3: When x<3, then x+3=(x+3), so in this case we'll have (x+3)=4x3 > x=0 > discard this value since 0 is not less than 3 (we consider the range when x<3). When x>=3, then x+3=x+3, so in this case we'll have x+3=4x3 > x=2 > this value of x is OK since 2>3. So, x + 3 = 4x  3 has only one root, x=2. Hope it's clear. Thanks Bunuel, this is how I solve the Modulus questions, so as a rule i guess, I should check both values by putting them back in eq, I use to think that they both are the roots since I considered both scenarios, ie one with X+3 and one with (x+3). I guess I will have to add this additional step in my methodology. Thanks.
_________________
Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!
1 Kudos = 1 thanks Nikhil



Intern
Status: Preparing...
Joined: 25 Mar 2013
Posts: 25
Location: United States
Sat: V
Concentration: Strategy, Technology
GMAT Date: 07222013
GPA: 3.7
WE: Information Technology (Computer Software)

Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
Show Tags
03 Jun 2013, 23:42
dvinoth86 wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos I generally think using graphs are faster for simple linear equations but you should be fast at plotting them as soon as you see the equations. For statement 2, if you plot the modulus graphs , you will see that one root is at x=0 and another root is some +ve x, 2 solutions hence insufficient i.e f(x) = x3 , f(0) = 3(y intercept) and x intercept is @ x=3 second f(x) = 2x 3 , f(0) = 3( y intercept is same as previous graph or solution 1) and x intercept is 3/2.



Senior Manager
Joined: 13 May 2013
Posts: 430

Re: [tricky one] Is x > 0? [#permalink]
Show Tags
04 Jun 2013, 07:56
Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > 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. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps.



Math Expert
Joined: 02 Sep 2009
Posts: 47079

Re: [tricky one] Is x > 0? [#permalink]
Show Tags
04 Jun 2013, 08:00
WholeLottaLove wrote: Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > 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. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. One can do this way too, but the way shown in my post is faster.
_________________
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: 13 May 2013
Posts: 430

Re: [tricky one] Is x > 0? [#permalink]
Show Tags
04 Jun 2013, 08:26
Hmmm... When I solve for the pos. and neg. values of x+3=4x3 I get: I. x+3=4x3 ==> 3x=6 ==> x=2II. x+3=(4x3) ==> x+3=4x+3 ==> 5x=0 ==> x=0So here in my presumably incorrect simplification, I have x=2 and x=0 in which case we can't be sure if x>0 For many abs. value questions it seems that you have to find the positive and negative cases each equation. I get that in this case, x+3=4x3 means that 4x3 is positive but why for other, similar questions, is solving for the positive and negative cases necessary? Bunuel wrote: WholeLottaLove wrote: Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > 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. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. One can do this way too, but the way shown in my post is faster.




Re: [tricky one] Is x > 0?
[#permalink]
04 Jun 2013, 08:26



Go to page
1 2 3
Next
[ 51 posts ]



