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For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 02:10
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 04:12
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Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Given ab = ba > This can only happen when ab<0 > b>a Also given a>0 , thus b>a>0 Now per question, a + 3 = a+3 as a >0 and thus a+3 >0 b = b as b>0 b – a = ba as b>a ab = ab as both a,b >0 Thus a + 3 + b + b – a + ab = a+3+b+ba+ab = 3+2b+ab. E is the correct answer.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 05:05
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Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Solution  Given that a – b = b – a, so b – a > 0, or b > a. Also given that a > 0, so b > 0. a + 3 + b + b – a + ab = a + 3 + b + b – a + ab =ab + 2b + 3 ANS E
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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Updated on: 22 Jul 2015, 05:51
Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Such Question require an intense observation of the given Information stepbystep Observation1: a – b = b – a which is possible only when signs of a and b are Same Since Given a > 0 so we figure out that a and b are both positive Observation2: a – b must be NonNegative and so should be the value of ba which is possible only when absolute value of b is greater than or equal to absolute value of a Now you may choose the values of a and b based on above observations e.g. b = 3 and a=1 and check the value of given functions and options a + 3 + b + b – a + ab = 1 + 3 + 3 + 3 – 1 + 1*3 = 12 A. ab + 3 = 1*3 + 3 = 0 not equal to 12 hence, INCORRECTB. ab + 3 = 1*3 + 3 = 6 not equal to 12 hence, INCORRECTC. ab + 2b + 3 = 1*3+2*3+3 = 6 not equal to 12 hence, INCORRECTD. ab + 2b – 2a – 3 = 1*3 + 2*3  2*1  3 = 4 not equal to 12 hence, INCORRECTE. ab + 2b + 3 = 1*3 + 2*3 + 3 = 12 CORRECT
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Originally posted by GMATinsight on 22 Jul 2015, 05:33.
Last edited by GMATinsight on 22 Jul 2015, 05:51, edited 1 time in total.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 05:39
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GMATinsight wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Such Question require an intense observation of the given Information stepbystep Observation1: a – b = b – a which is possible only when signs of a and b are Same Also, Since Given a > 0 so we figure out that a and b are both positive Observation2: a – b must be NonNegative and so should be the value of ba which is possible only when absolute value of b is greater than or equal to absolute value of a Now you may choose the values of a and b based on above observations e.g. b = 3 and a=1 and check the value of given functions and options a + 3 + b + b – a + ab = 1 + 3 + 3 + 3 – 1 + 1*3 = 12 GMATinsight, the text in red above is not necessarily true. ab = ba ONLY when ab <0 or b>a . now b > a by being (3,1) (opposite signs) or by (4,2) or by (3,1) (in these 2 cases, same signs)



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 05:50
Engr2012 wrote: GMATinsight wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Such Question require an intense observation of the given Information stepbystep Observation1: a – b = b – a which is possible only when signs of a and b are Same Also, Since Given a > 0 so we figure out that a and b are both positive Observation2: a – b must be NonNegative and so should be the value of ba which is possible only when absolute value of b is greater than or equal to absolute value of a Now you may choose the values of a and b based on above observations e.g. b = 3 and a=1 and check the value of given functions and options a + 3 + b + b – a + ab = 1 + 3 + 3 + 3 – 1 + 1*3 = 12 GMATinsight, the text in red above is not necessarily true. ab = ba ONLY when ab <0 or b>a . now b > a by being (3,1) (opposite signs) or by (2,4) or by (3,1) (in these 2 cases, same signs) a and b will always have same sign for a – b = b – a to be true for given a >0The word "also" was disturbing the meaning probably, so removed it. and b may have opposite sign only when a is negative which was out of question due to given information a>0
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 06:31
GMATinsight wrote: GMATinsight, the text in red above is not necessarily true. ab = ba ONLY when ab <0 or b>a . now b > a by being (3,1) (opposite signs) or by (2,4) or by (3,1) (in these 2 cases, same signs) a and b will always have same sign for a – b = b – a to be true for given a >0The word "also" was disturbing the meaning probably, so removed it. and b may have opposite sign only when a is negative which was out of question due to given information a>0 Yes, exactly. To me, you were mentioning about a>0 as an additional information after the statement. This seemed incorrect. Thanks for the clarification.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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22 Jul 2015, 12:52
Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. ab = (ab), which means that ab is <=0. Since a>0, then b>a>0. let a = 1 and b = 2 a + 3 + b + b – a + ab = 4+2+1+2 = 9 Sub the value of a and b in answer options E. 9 Hence Option E



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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23 Jul 2015, 07:22
Engr2012 wrote: GMATinsight wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Such Question require an intense observation of the given Information stepbystep Observation1: a – b = b – a which is possible only when signs of a and b are Same Also, Since Given a > 0 so we figure out that a and b are both positive Observation2: a – b must be NonNegative and so should be the value of ba which is possible only when absolute value of b is greater than or equal to absolute value of a Now you may choose the values of a and b based on above observations e.g. b = 3 and a=1 and check the value of given functions and options a + 3 + b + b – a + ab = 1 + 3 + 3 + 3 – 1 + 1*3 = 12 GMATinsight, the text in red above is not necessarily true. ab = ba ONLY when ab <0 or b>a . now b > a by being (3,1) (opposite signs) or by (2,4) or by (3,1) (in these 2 cases, same signs) Hi Engr2012, I can understand that both should have the same signs. But I think the example of negative signs should be revered. i.e. (2,4) to (4,2) to satisfy that b>a as 2>4. I think there is a typo.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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26 Jul 2015, 12:12



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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27 Jul 2015, 10:43
Mo2men wrote: Hi Engr2012, I can understand that both should have the same signs. But I think the example of negative signs should be revered. i.e. (2,4) to (4,2) to satisfy that b>a as 2>4. I think there is a typo.
Yes, thanks. It was a typo.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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27 Jul 2015, 17:37
Hi All, While this prompt looks complex, it can be solved rather easily by TESTing VALUES. We're told that AB = BA. While there are LOTS of values that will fit this equation, the easiest 'option' is to make A=B. We're also told that A > 0. IF... A=1 B=1 We're asked for the value of A+3 + B + BA + AB... 1+3 +1 + 0 + 1 = 6 Answer A: AB + 3 = 2 NOT a match Answer B: AB + 3 = 4 NOT a match Answer C: AB + 2B + 3 = 4 NOT a match Answer D: AB + 2B – 2A – 3 = 2 NOT a match Answer E: AB + 2B + 3 = 6 This IS a MATCH Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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27 Jul 2015, 17:43
Bunuel wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. 800score Official Solution:We know that a – b = b – a, so b – a > 0, or b > a. We also know that a > 0, so b > 0. Knowing that both a and b are positive we can easily simplify: a + 3 + b + b – a + ab = a + 3 + b + b – a + ab = ab + 2b + 3 The right answer is choice (E).Hi All, This explanation is not completely correct. A and B can be the SAME value. Be mindful of these types of Number Property details that you might miss  while it would not impact your solution to this particular question, it could cost you points in other areas (especially in DS). GMAT assassins aren't born, they're made, Rich
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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27 Jul 2015, 17:49
EMPOWERgmatRichC wrote: Bunuel wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. 800score Official Solution:We know that a – b = b – a, so b – a > 0, or b > a. We also know that a > 0, so b > 0. Knowing that both a and b are positive we can easily simplify: a + 3 + b + b – a + ab = a + 3 + b + b – a + ab = ab + 2b + 3 The right answer is choice (E).Hi All, This explanation is not completely correct. A and B can be the SAME value. Be mindful of these types of Number Property details that you might miss  while it would not impact your solution to this particular question, it could cost you points in other areas (especially in DS). GMAT assassins aren't born, they're made, Rich Hi EMPOWERgmatRichCThank you for your solution. But I have a question about your statement "A and B can be the SAME value". IMHO, this is not true. We have been given that a – b = (ab) . This is similar to x = x for all x<0. We do not include equality with the negative value. Please clarify.



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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27 Jul 2015, 18:02
Hi Engr2012, The easiest way to prove the point is to TEST values for A and B that are the SAME.... You can either use physical values or substitute in a new variable.... IF... A=B=2 AB = BA 22 = 0 22 = 0 0 = 0 Thus, we have proof that A CAN = B Or, using a variable.... A=B=X XX = XX XX = 0 XX = 0 0 = 0 Again, proof that A CAN = B GMAT assassins aren't born, they're made, Rich
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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07 Jul 2017, 06:07
Hi balamoon, I have a ques on the step below: Solution  Given that a – b = b – a, so b – a > 0, or b > a. Also given that a > 0, so b > 0. a + 3 + b + b – a + ab = a + 3 + b + b – a + ab =ab + 2b + 3 here b becomes just b. How did that happen? The rest of the modulus openings retained thier signs but here it changed. Can anyone explain how?
Thank you !ANS E



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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08 Aug 2017, 07:58
balamoon wrote: Bunuel wrote: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + b + b – a + ab =
A. ab + 3 B. ab + 3 C. ab + 2b + 3 D. ab + 2b – 2a – 3 E. ab + 2b + 3
Kudos for a correct solution. Solution  Given that a – b = b – a, so b – a > 0, or b > a. Also given that a > 0, so b > 0. a + 3 + b + b – a + ab = a + 3 + b + b – a + ab =ab + 2b + 3 ANS E why ba>0 Please help me to understand



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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08 Aug 2017, 08:49
Madhavi1990 wrote: Hi balamoon, I have a ques on the step below: Solution  Given that a – b = b – a, so b – a > 0, or b > a. Also given that a > 0, so b > 0. a + 3 + b + b – a + ab = a + 3 + b + b – a + ab =ab + 2b + 3 here b becomes just b. How did that happen? The rest of the modulus openings retained thier signs but here it changed. Can anyone explain how?
Thank you !ANS E This is because b>a>0; If b = 3, b = 3 & 3= 3



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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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13 Sep 2017, 10:41
I interpreted it like this: ab= ba This can also be written as: ba= ba That means ba is positive. (a>0) ba>0 b>0 Rest is same.
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Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 + [#permalink]
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13 Oct 2017, 09:41
Bunuel wrote: For any a and b that satisfy \(a – b = b – a\) and \(a > 0\), then \(a + 3 + b + b – a + ab =\)
A. \(ab + 3\)
B. \(ab + 3\)
C. \(ab + 2b + 3\)
D. \(ab + 2b – 2a – 3\)
E. \(ab + 2b + 3\)
Kudos for a correct solution. Given ab = b  a b>a, for ab = positive a>0 therefore, a+3>0 Let a = 1, b=2 4+ 2 + 1 + 2 = 9 ab + 2b + 3 = 2+4+3 = 9 Therefore, (E) ab + 2b + 3




Re: For any a and b that satisfy a – b = b – a and a > 0, then a + 3 +
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