Find all School-related info fast with the new
Applicant profile
My workspace
GMAT Courses
Private Tutoring
Admissions Consulting
Top 20 by # of applicants
Not a Member?
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.
Customizedfor You
we will pick new questions that match your level based on your Timer History
TrackYour Progress
every week, we’ll send you an estimated GMAT score based on your performance
PracticePays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem,
Hello Guest!
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
and many more benefits!
Find us on Facebook
E-mail address:
Confirm password:
CREATEAN ACCOUNT
We’ll give you an estimate of your score
We’ll provide personalized question recommendations
Your score will improve and your results will be more realistic
Events & Promotions
Events & Promotions in June
CLICK HERE TO HIDE/SHOW EVENTS
CLICK HERE TO HIDE/SHOW EVENTS
Joined: 19 Oct 2011
Posts: 132
Location: India
Followers: 3
, given: 33
This post receivedKUDOS
This post wasBOOKMARKED
Question Stats:
42% (02:25) correct
58% (01:29) wrong
based on 1069 sessions
Is x & 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|Liked the question? encourage by giving kudos
Spoiler: OA
_________________
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&/forum/the-one-thing-you-wish-you-knew-gmat-club-contest-140358.html#p1130989
Math Expert
Joined: 02 Sep 2009
Posts: 34159
Followers: 6132
, given: 10002
This post receivedKUDOS
Expert's post
This post wasBOOKMARKED
Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), 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) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.
_________________
New to the Math Forum? Please read this:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Joined: 29 Aug 2011
Followers: 1
, given: 3
This post receivedKUDOS
This post wasBOOKMARKED
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|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(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: 34159
Followers: 6132
, given: 10002
This post receivedKUDOS
Expert's post
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|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(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.RegardsIt seems that you understand this method very well. Everything is correct except the red parts: -(x-3) = -(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:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Joined: 29 Aug 2011
Followers: 1
[], given: 3
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|
|x -3| = x -3 for x-3 &= 0 ---& x &= 3
|x -3| = -(x -3) for x-3 & 0 ---& x & 3
|2x -3 | = 2x -3 for 2x -3 &= 0 ----& x &=3/2
|2x -3 | = -(2x -3) for 2x -3 & 0 ----& x & 3/2So we have 3 ranges 1) x &=3 ==& x - 3 = 2x -3 ====& x
= 0 ; discard2) 3/2 & x & 3 ===& -(x-3) = 2x -3 ====& -x + 3 = 2x -3 ===& x =2 ( accepted solution )3) x & 3/2 ===& -(x-3) = -(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.RegardsIt seems that you understand this method very well. Everything is correct except the red parts: -(x-3) = -(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: 34159
Followers: 6132
, given: 10002
This post receivedKUDOS
Expert's post
This post wasBOOKMARKED
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.RegardsA. 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 can not square both sides and write \(x^2&y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.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 non-negative we can safely apply squaring.Hope it helps.
_________________
New to the Math Forum? Please read this:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Joined: 26 Jan 2012
Followers: 0
[], given: 0
Well, I have a little bit another solution for this problem Consider (1): if x&-3 then x+3=4x-3, so x=2, sufficient if x&-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x&-3Sum 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 sufficientIn your logic, you gave a mistake, because you've missed the root. Posted from
Math Expert
Joined: 02 Sep 2009
Posts: 34159
Followers: 6132
[], given: 10002
Expert's post
terance wrote:Well, I have a little bit another solution for this problem Consider (1): if x&-3 then x+3=4x-3, so x=2, sufficient if x&-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x&-3Sum 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 sufficientIn your logic, you gave a mistake, because you've missed the root. Posted from Welcome 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:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Joined: 27 Oct 2011
Posts: 191
Location: United States
Concentration: Finance, Strategy
Schools: , , , ,
WE: Account Management (Consumer Products)
Followers: 4
[], given: 4
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!!!
Senior Manager
Joined: 30 Jun 2011
Posts: 274
Followers: 0
[], given: 20
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 .. INVALIDHence Sufficient(2) |x – 3| = |2x – 3| =& x = 0 or 6 .. INVALIDHence In-Sufficient
Joined: 09 Apr 2012
Followers: 0
[], given: 3
Is this a correct way to approach 2) |x – 3| = |2x – 3| ?Step 1:(x-3) = (2x-3)=& x = 0Substitute x = 0 in |x – 3| = |2x – 3| =& VALIDStep 2:(x-3) = -(2x-3)=& x = 2Substitute x = 2 in |x – 3| = |2x – 3| =& VALIDSo 2) |x – 3| = |2x – 3| is NOT SUFFICIENT
Senior Manager
Joined: 30 Jun 2011
Posts: 274
Followers: 0
[], given: 20
kirncha wrote:Is this a correct way to approach 2) |x – 3| = |2x – 3| ?Step 1:(x-3) = (2x-3)=& x = 0Substitute x = 0 in |x – 3| = |2x – 3| =& VALIDStep 2:(x-3) = -(2x-3)=& x = 2Substitute x = 2 in |x – 3| = |2x – 3| =& VALIDSo 2) |x – 3| = |2x – 3| is NOT SUFFICIENTYes but you can skip the substitution part, read my statement above
Joined: 02 Apr 2012
Followers: 0
[], given: 3
Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), 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) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.Pretty slick answer. Saves a lot of time.
Joined: 04 Dec 2011
Followers: 0
[], given: 13
Hi BunuelThis is how I solved the 2 statements, and 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 thanksNikhil
Math Expert
Joined: 02 Sep 2009
Posts: 34159
Followers: 6132
, given: 10002
This post receivedKUDOS
Expert's post
nikhil007 wrote:Hi BunuelThis is how I solved the 2 statements, and 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)=4x-3 --& 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=4x-3 --& 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:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Joined: 04 Dec 2011
Followers: 0
, given: 13
This post receivedKUDOS
Bunuel wrote:nikhil007 wrote:Hi BunuelThis is how I solved the 2 statements, and 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)=4x-3 --& 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=4x-3 --& 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 thanksNikhil
Status: Preparing...
Joined: 25 Mar 2013
Location: United States
Concentration: Strategy, Technology
GMAT Date: 07-22-2013
WE: Information Technology (Computer Software)
Followers: 0
[], given: 14
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 insufficienti.e f(x) = x-3 , |f(0)| = 3(y intercept) and x intercept is @ x=3second 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: 472
Followers: 3
[], given: 134
Hi!Why do we not find the positive and negative values for |x+3|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), 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) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) or \(x=2\). Not sufficient.Answer: A.Hope it helps.
Math Expert
Joined: 02 Sep 2009
Posts: 34159
Followers: 6132
, given: 10002
This post receivedKUDOS
Expert's post
WholeLottaLove wrote:Hi!Why do we not find the positive and negative values for |x+3|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), 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) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) 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:
| PLEASE READ AND FOLLOW:
Resources:
Collection of Questions:PS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ; 9 ; 10
DS: 1. ; 2. ; 3. ; 4. ; 5. ; 6. ; 7 ; 8 ;
9 ; 10 , 11
Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 3
[], given: 134
Hmmm...When I solve for the pos. and neg. values of |x+3|=4x-3 I get:I. x+3=4x-3
x=2II. x+3=-(4x-3)
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&0For 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|=4x-3 means that 4x-3 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|=4x-3?
Is it because this isn't a &, &, &= problem?Thanks! Bunuel wrote:Is x & 0?(1) \(|x+3|=4x-3\) --& LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --& \(4x-3\geq{0}\) --& \(x\geq{\frac{3}{4}}\), 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) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --&
\((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --& \(x(3x-6)=0\) --& \(x=0\) 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.
gmatclubot
Similar topics
SimilarTopics:
Display posts from previous:
All posts1 day7 days2 weeks1 month3 months6 months1 year
AuthorPost timeSubject
AscendingDescending
Moderators:
&31&by Bunuel&&30&by Bunuel&&17&by rohitgoel15&&13&by Bunuel&&13
DS Expert Advice
Subscribe to RSS headline updates from:
Powered by FeedBurner
Latest Kudos
&by lylya4 &
&by muralimba &
&by hemanthp &
&by tripathymadhu31 &
&by Bunuel &
&by avohden &
&by zerotoinfinite2006 &
&by Bunuel &
You are here:
GMAT (R) is a registered trademark of the Graduate Management Admission Council (R) (GMAC (R)).
GMAT Club's website has not been reviewed or endorsed by GMAC.
GMAT Resources
MBA Resources
Copyright & GMAT Club
GMAT & is a registered trademark of the Graduate Management Admission Council & (GMAC &). GMAT Club's website has not been reviewed or endorsed by GMAC.
& DeeP 2016
Web Design & Development
The post is bookmarked successfully
by GMAT Data Sufficiency (DS)