GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Aug 2018, 10:54

Close

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.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

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, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47945
For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 22 Jul 2015, 02:10
5
34
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

67% (01:38) correct 33% (01:39) wrong based on 837 sessions

HideShow timer Statistics

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.

_________________

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

Most Helpful Community Reply
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 22 Jul 2015, 04:12
6
9
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.
General Discussion
Manager
Manager
User avatar
Joined: 26 Dec 2011
Posts: 116
Schools: HBS '18, IIMA
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 22 Jul 2015, 05:05
6
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
_________________

Thanks,
Kudos Please

SVP
SVP
User avatar
P
Joined: 08 Jul 2010
Posts: 2137
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post Updated on: 22 Jul 2015, 05:51
1
1
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
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION


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.
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 22 Jul 2015, 05:39
1
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)
SVP
SVP
User avatar
P
Joined: 08 Jul 2010
Posts: 2137
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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.
Manager
Manager
avatar
Joined: 20 Jul 2011
Posts: 80
GMAT 1: 660 Q49 V31
GMAT ToolKit User
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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
SVP
SVP
User avatar
D
Joined: 26 Mar 2013
Posts: 1777
Reviews Badge CAT Tests
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47945
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 26 Jul 2015, 12:12
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

Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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.
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12181
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 27 Jul 2015, 17:37
1
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:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12181
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 27 Jul 2015, 17:43
1
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
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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.
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12181
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Senior Manager
Senior Manager
avatar
S
Joined: 15 Jan 2017
Posts: 367
CAT Tests
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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
Intern
Intern
avatar
B
Joined: 13 Jul 2017
Posts: 29
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

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
Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 734
Premium Member
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 13 Sep 2017, 10:41
I interpreted it like this:

|a-b|= b-a

This can also be written as:
|b-a|= b-a

That means b-a is positive. (a>0)
b-a>0
b>0

Rest is same.
_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.

Manager
Manager
User avatar
B
Joined: 21 Jun 2017
Posts: 84
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

Show Tags

New post 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 |a-b| = b - a
b>a, for |a-b| = 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| + &nbs [#permalink] 13 Oct 2017, 09:41

Go to page    1   2    Next  [ 21 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.