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

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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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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.
[Reveal] Spoiler: OA

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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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New post 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 |a-b| = b-a ---> This can only happen when a-b<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| = b-a as b>a
|ab| = ab as both a,b >0

Thus |a + 3| + |-b| + |b – a| + |ab| = a+3+b+b-a+ab = 3+2b+ab. E is the correct answer.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
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Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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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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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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For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| + [#permalink]

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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.


Such Question require an intense observation of the given Information step-by-step

Observation-1: |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

Observation-2: |a – b| must be Non-Negative and so should be the value of b-a 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, INCORRECT
B. ab + 3 = 1*3 + 3 = 6 not equal to 12 hence, INCORRECT
C. -ab + 2b + 3 = -1*3+2*3+3 = 6 not equal to 12 hence, INCORRECT
D. ab + 2b – 2a – 3 = 1*3 + 2*3 - 2*1 - 3 = 4 not equal to 12 hence, INCORRECT
E. ab + 2b + 3 = 1*3 + 2*3 + 3 = 12 CORRECT
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Last edited by GMATinsight on 22 Jul 2015, 05:51, edited 1 time in total.

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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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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 step-by-step

Observation-1: |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

Observation-2: |a – b| must be Non-Negative and so should be the value of b-a 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.

|a-b| = b-a ONLY when a-b <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)
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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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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New post 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 step-by-step

Observation-1: |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

Observation-2: |a – b| must be Non-Negative and so should be the value of b-a 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.

|a-b| = b-a ONLY when a-b <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 >0

The 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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Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

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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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New post 22 Jul 2015, 06:31
GMATinsight wrote:

GMATinsight, the text in red above is not necessarily true.

|a-b| = b-a ONLY when a-b <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 >0

The 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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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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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New post 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.


|a-b| = -(a-b), which means that a-b 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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New post 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 step-by-step

Observation-1: |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

Observation-2: |a – b| must be Non-Negative and so should be the value of b-a 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.

|a-b| = b-a ONLY when a-b <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.

Kudos [?]: 262 [0], given: 154

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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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New post 26 Jul 2015, 12:12
Expert's post
1
This post was
BOOKMARKED
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).
_________________

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?
Extra-hard Quant Tests with Brilliant Analytics

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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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New post 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.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

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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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New post 27 Jul 2015, 17:37
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Hi All,

While this prompt looks complex, it can be solved rather easily by TESTing VALUES.

We're told that |A-B| = B-A. 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| + |B-A| + |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:
[Reveal] Spoiler:
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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New post 27 Jul 2015, 17:43
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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).

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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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New post 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 EMPOWERgmatRichC

Thank 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| = -(a-b) . 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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New post 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

|A-B| = B-A
|2-2| = 0
2-2 = 0
0 = 0

Thus, we have proof that A CAN = B

Or, using a variable....
A=B=X

|X-X| = X-X
|X-X| = 0
X-X = 0
0 = 0

Again, proof that A CAN = B

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

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New post 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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New post 03 Aug 2017, 23:49
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



same doubt, please someone explain!
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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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New post 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 b-a>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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New post 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] 08 Aug 2017, 08:49

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