Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

Is x > 0? (1) x + 3 = 4x 3 (2) x + 1 = 2x 1 Can [#permalink]
Show Tags
03 Sep 2010, 14:20
2
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
62% (02:14) correct
38% (01:26) wrong based on 288 sessions
HideShow timer Statistics
Is x > 0? (1) x + 3 = 4x – 3 (2) x + 1 = 2x – 1 Can someone explain this to me
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If the Q jogged your mind do Kudos me : )



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: Absolute value DS [#permalink]
Show Tags
03 Sep 2010, 14:29
6
This post received KUDOS
Expert's post
3
This post was BOOKMARKED



Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

Re: Absolute value DS [#permalink]
Show Tags
03 Sep 2010, 14:43
1
This post received KUDOS
1
This post was BOOKMARKED
thanks that helped your explanation is more simpler and fundamental! the below explanation i did not get especially the last part where we need to substitute and check (1) SUFFICIENT: Here, we are told that x + 3 = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are: x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0 However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them. (2) + 3 = 4(2) – 3 and (0) + 3 = 4(0) – 3 5 = 8 – 3 3 = –3 5 = 5 3 = –3 Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive.
_________________
If the Q jogged your mind do Kudos me : )



Senior Manager
Joined: 20 Jul 2010
Posts: 263

Re: Absolute value DS [#permalink]
Show Tags
20 Sep 2010, 19:44
1
This post received KUDOS
amitjash wrote: Whenever i see this type of question i start to solve for X. Now in this case, from st1 i got X = 2 or 0 From st2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check. I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems?
_________________
If you like my post, consider giving me some KUDOS !!!!! Like you I need them



Retired Moderator
Joined: 02 Sep 2010
Posts: 803
Location: London

Re: Absolute value DS [#permalink]
Show Tags
20 Sep 2010, 20:10
1
This post received KUDOS
saxenashobhit wrote: amitjash wrote: Whenever i see this type of question i start to solve for X. Now in this case, from st1 i got X = 2 or 0 From st2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check. I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems? When you are dealing with absolute values, and you solve it in 2 cases assuming first that expression >= 0 and then that expression < 0, you must verify the solution you get to make sure you don't violate the extra assumption you made in the beginning Eg.When you assume x+3>0, it means you assume x>3. You get an answer 2, which is valid as no assumptions are violated When you assume x+3<0, it means you assume x<3. You get an answer 0, which is invalid given your assumption You do not need to necessarily "plug the solution back in", it is sufficient to check against the assumption you made while deciding the sign of the absolute value
_________________
Math writeups 1) Algebra101 2) Sequences 3) Set combinatorics 4) 3D geometry
My GMAT story
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: Absolute value DS [#permalink]
Show Tags
20 Sep 2010, 21:54
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
saxenashobhit wrote: amitjash wrote: Whenever i see this type of question i start to solve for X. Now in this case, from st1 i got X = 2 or 0 From st2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check. I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems? Solving for \(x\) is not the best way to deal with this problem. But if you do this way you should know that checking validity of the solutions is very important for absolute value questions. When you solve these kind of questions with check point method (discussed at: someinequalitiesquestions93760.html?hilit=substitute%20method), you test validity of solution on the stage of obtaining these values (by checking whether the value is in the range you are testing at the moment) and if the value obtained IS in the range you are testing at the moment, you don't need to substitute it afterwards to check, you've already done the checking and if the value obtained IS NOT in the range you are testing at the moment you also don't need to substitute it afterwards to check, you've already done the checking. When you just expand the absolute value once with negative sign and once with positive sign then you should check whether obtained solution(s) satisfy equation be substituting solution(s) back to the equation. Check 3steps approach in Wlaker's post on Absolute Value for more on this: mathabsolutevaluemodulus86462.htmlHope it helps.
_________________
New to the Math Forum? Please read this: 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



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: Absolute value DS [#permalink]
Show Tags
03 Sep 2010, 16:15
rxs0005 wrote: thanks that helped your explanation is more simpler and fundamental!
the below explanation i did not get especially the last part where we need to substitute and check
(1) SUFFICIENT: Here, we are told that x + 3 = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:
x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0
However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.
(2) + 3 = 4(2) – 3 and (0) + 3 = 4(0) – 3 5 = 8 – 3 3 = –3 5 = 5 3 = –3
Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive. This issue is discussed here: someinequalitiesquestions93760.html?hilit=substitute%20method
_________________
New to the Math Forum? Please read this: 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: 17 Mar 2010
Posts: 183

Re: Absolute value DS [#permalink]
Show Tags
07 Sep 2010, 23:33
Whenever i see this type of question i start to solve for X. Now in this case, from st1 i got X = 2 or 0 From st2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.



Manager
Joined: 30 May 2010
Posts: 190

Re: Absolute value DS [#permalink]
Show Tags
20 Sep 2010, 00:42
rxs0005 wrote: thanks that helped your explanation is more simpler and fundamental!
the below explanation i did not get especially the last part where we need to substitute and check
(1) SUFFICIENT: Here, we are told that x + 3 = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:
x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0
However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.
(2) + 3 = 4(2) – 3 and (0) + 3 = 4(0) – 3 5 = 8 – 3 3 = –3 5 = 5 3 = –3
Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive. This appears every so often on absolute value questions. It relies on you not testing the possible solution. You must always verify the solutions you get in your first step. I've made this mistake several times



Manager
Joined: 04 Aug 2010
Posts: 154

Re: Absolute value DS [#permalink]
Show Tags
20 Sep 2010, 19:18
Bunuel wrote: (1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative
great way to look at it. thanks for the tip



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15982

Re: Is x > 0? (1) x + 3 = 4x 3 (2) x + 1 = 2x 1 Can [#permalink]
Show Tags
08 Oct 2013, 17:17
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



Intern
Joined: 26 Feb 2012
Posts: 16
Concentration: Strategy, International Business

Re: Absolute value DS [#permalink]
Show Tags
09 Oct 2013, 07:28
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.
(2) \(x+1=2x1\) > the same here > \(2x1\geq{0}\) > \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.
So you see that you don't even need to find exact value(s) of \(x\) to answer the question.
Answer: D.
Hope it helps. For those more inclined to solving equations, here is a different approach: (1) x+3=4x3 If x+3>0 then: a) x > 3 AND b) x+3 = 4x3 => 3x=6 => x=2. Since x=2 does NOT contradict x>3, x=2 is a solution and x is positive However if x+3<0 then: a) x<3 AND b) x+3=4x+3 => x=0. BUT x must be less than 3, so x=0 is NOT a solution. Therefore only the case where x+3>0 can be possible being x=2 the only possible solution, therefore always positive. (2) x+1=2x1 Works exactly the same: If x+1>0 then: a) x > 1 AND b) x+1 = 2x1 => x = 2 Since x=2 does NOT contradict x>1, x=2 is a solution and x is positive However if x+1<0 then: a) x<1 AND b) x+1 = 2x+1 => x = 0. BUT x must be less than 1, so x=0 is NOT a solution. Therefore only the case where x+1>0 can be possible being x=2 the only possible solution, therefore always positive. In BOTH cases the only possible value for x is always positive. Therefore D is the answer.



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15982

Re: Is x > 0? (1) x + 3 = 4x 3 (2) x + 1 = 2x 1 Can [#permalink]
Show Tags
02 Jan 2015, 07:07
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



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15982

Re: Is x > 0? (1) x + 3 = 4x 3 (2) x + 1 = 2x 1 Can [#permalink]
Show Tags
17 Mar 2016, 01:40
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? (1) x + 3 = 4x 3 (2) x + 1 = 2x 1 Can
[#permalink]
17 Mar 2016, 01:40







