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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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29 Dec 2015, 23:12
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution Is x > 0? (1) x + 3 < 4 (2) x  3 < 4 In case of inequality questions, it is important to note that conditions are sufficient if the range of the question includes the range of the conditions. There is 1 variable (x) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance that D is the answer. In case of the condition 1), we can obtain 4<x+3<4. This yields 7<x<1. Since the range of the question does not include the range of the condition, the condition 1) is not sufficient. In case of the condition 2), we can obtain 4<x3<4. This also yields, 1<x<7. Since the range of the question does not include the range of the condition, the condition 2) is not sufficient. Using both the condition 1) and 2), since 1<x<1, the range of the question does not include the range of the conditions. Therefore, the conditions are not sufficient and the correct answer is E. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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18 Aug 2016, 07:31
MarkusKarl wrote: 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. Hi Bunuel! Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about? Best wishes When there are absolute values on both sides you can safely square. When there is an absolute value on one side and nonabsolute value on another it becomes trickier  you should plug back the solutions to make sure that they are valid. You can search for similar questions here: search.php?search_id=tag&tag_id=37Hope it helps.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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22 Dec 2016, 19:55
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. Thanks Bunuel  this was such a quick & efficient way of solving this problem.



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 Dec 2017, 10:30
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. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, x3=2x3, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 Dec 2017, 10:34
bapoon1991 wrote: 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. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, x3=2x3, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev You would have been right if it were \(x3=x3/2\) but it's \(x3=2*x3/2\). So, the distance from x to 3 is TWICE the distance from x to 3/2.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 Dec 2017, 10:41
Bunuel wrote: bapoon1991 wrote: 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. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, x3=2x3, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev You would have been right if it were \(x3=x3/2\) but it's \(x3=2*x3/2\). So, the distance from x to 3 is TWICE the distance from x to 3/2. Thanks a lot for your super fast response! After your explanation, everything is clear now. So thanks again! Lastly, I have my GMAT in the next 8 days, and I am facing a bit of an issue solving 700+ DS questions. Hence, it will be great if you can share with me any link, PDF which has a group of 700+ DS questions which I can practice before my Exam. Thanks, Dev



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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15 Jul 2018, 06:37
Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution )[Did not understand, Why ve sign is here as the x range is positive]3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.
What am i missing here and how to solve these questions effectively Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). SOLUTION:We have two transition points for \(x3=2x3\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\) Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When \(x<\frac{3}{2}\), then \(x3\) is negative and \(2x3\) is negative too, thus \(x3=(x3)\) and \(2x3=(2x3)\). Therefore for this range \(x3=2x3\) transforms to \((x3)=(2x3)\) > \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)). 2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x3\) is negative and \(2x3\) is positive, thus \(x3=(x3)\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \((x3)=2x3\) > \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)). 3. When \(3<x\), then \(x3\) is positive and \(2x3\) is positive too, thus \(x3=x3\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \(x3=2x3\) > \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)). Thus \(x3=2x3\) has two solutions \(x=0\) and \(x=2\). Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/isx01x3 ... l#p1048512Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x  3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x  3, it will result in negative number, however if i plug in 3 in x  3, it will result in 0 or nonnegative, and therefore x  3 can be negative or nonnegative. In your post x  3 must be negative, how do we eliminate the nonnegative one ? Or do we simply ignore nonnegative ?



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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15 Jul 2018, 07:06
tixan wrote: Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution )[Did not understand, Why ve sign is here as the x range is positive]3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.
What am i missing here and how to solve these questions effectively Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). SOLUTION:We have two transition points for \(x3=2x3\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\) Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When \(x<\frac{3}{2}\), then \(x3\) is negative and \(2x3\) is negative too, thus \(x3=(x3)\) and \(2x3=(2x3)\). Therefore for this range \(x3=2x3\) transforms to \((x3)=(2x3)\) > \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)). 2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x3\) is negative and \(2x3\) is positive, thus \(x3=(x3)\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \((x3)=2x3\) > \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)). 3. When \(3<x\), then \(x3\) is positive and \(2x3\) is positive too, thus \(x3=x3\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \(x3=2x3\) > \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)). Thus \(x3=2x3\) has two solutions \(x=0\) and \(x=2\). Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/isx01x3 ... l#p1048512Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x  3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x  3, it will result in negative number, however if i plug in 3 in x  3, it will result in 0 or nonnegative, and therefore x  3 can be negative or nonnegative. In your post x  3 must be negative, how do we eliminate the nonnegative one ? Or do we simply ignore nonnegative ? Yes, \({x}\leq{3}\), gives \(x  3 \leq 0\) but \(x3=(x3)\) is still true. That's because a = a when a <= 0. Meaning that if a is negative or 0, then a = a. For example, if say a = 1, then a = a = (1) = 1 and if say a = 0, then a = a = 0 = 0. Does this make sense?
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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.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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17 Jul 2018, 05:34
Bunuel wrote: tixan wrote: Bunuel wrote: Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). SOLUTION:We have two transition points for \(x3=2x3\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\) Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When \(x<\frac{3}{2}\), then \(x3\) is negative and \(2x3\) is negative too, thus \(x3=(x3)\) and \(2x3=(2x3)\). Therefore for this range \(x3=2x3\) transforms to \((x3)=(2x3)\) > \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)). 2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x3\) is negative and \(2x3\) is positive, thus \(x3=(x3)\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \((x3)=2x3\) > \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)). 3. When \(3<x\), then \(x3\) is positive and \(2x3\) is positive too, thus \(x3=x3\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \(x3=2x3\) > \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)). Thus \(x3=2x3\) has two solutions \(x=0\) and \(x=2\). Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/isx01x3 ... l#p1048512 Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x  3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x  3, it will result in negative number, however if i plug in 3 in x  3, it will result in 0 or nonnegative, and therefore x  3 can be negative or nonnegative. In your post x  3 must be negative, how do we eliminate the nonnegative one ? Or do we simply ignore nonnegative ? Yes, \({x}\leq{3}\), gives \(x  3 \leq 0\) but \(x3=(x3)\) is still true. That's because a = a when a <= 0. Meaning that if a is negative or 0, then a = a. For example, if say a = 1, then a = a = (1) = 1 and if say a = 0, then a = a = 0 = 0. Does this make sense? Dear Bunuel  yes, it makes sense and thanks very very much for the explanation!! it is very clear now!



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Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 01:32
dvinoth86 wrote: Is x > 0?
(1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
(1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 10:13
Mo2men wrote: dvinoth86 wrote: Is x > 0?
(1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
(1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x3 and 2x3) do NOT have to be nonnegative. One solution for Statement 2 is x=0, with the result that x3 = 3 and 2x3 = 3.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 13:08
GMATGuruNY wrote: Mo2men wrote: (1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x3 and 2x3) do NOT have to be nonnegative. One solution for Statement 2 is x=0, with the result that x3 = 3 and 2x3 = 3. Thanks GMATGuruNY. In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign?



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 23:07
Mo2men wrote: Thanks GMATGuruNY. In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign? Since the question stem asks whether x>0  and an equation with absolute value on both sides can have both a positive and a nonpositive solution  I recommend that you solve the equation, either by opening up the modulus or by squaring both sides.
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