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

 It is currently 20 Feb 2019, 12:31

### 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

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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST
• ### Online GMAT boot camp for FREE

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# 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
Joined: 02 Sep 2009
Posts: 53020
For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 01:10
5
42
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:13) correct 34% (02:09) wrong based on 929 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.

_________________
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 03:12
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
Joined: 26 Dec 2011
Posts: 114
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

22 Jul 2015, 04:05
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,

CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2788
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

Updated on: 22 Jul 2015, 04:51
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

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

Originally posted by GMATinsight on 22 Jul 2015, 04:33.
Last edited by GMATinsight on 22 Jul 2015, 04:51, edited 1 time in total.
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 04:39
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)
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2788
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 04:50
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

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

CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 05:31
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
Joined: 20 Jul 2011
Posts: 80
GMAT 1: 660 Q49 V31
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

22 Jul 2015, 11:52
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.

|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
Joined: 26 Mar 2013
Posts: 2067
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

23 Jul 2015, 06:22
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
Joined: 02 Sep 2009
Posts: 53020
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

26 Jul 2015, 11:12
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).
_________________
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

27 Jul 2015, 09:43
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
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13562
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

27 Jul 2015, 16:37
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

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 Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13562 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 27 Jul 2015, 16:43 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!*****

CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

27 Jul 2015, 16:49
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
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13562
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

27 Jul 2015, 17:02
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
Joined: 15 Jan 2017
Posts: 353
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

07 Jul 2017, 05:07
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
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

08 Aug 2017, 06:58
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
CEO
Joined: 20 Mar 2014
Posts: 2629
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

08 Aug 2017, 07:49
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
Joined: 02 Sep 2016
Posts: 670
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

13 Sep 2017, 09:41
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
Joined: 21 Jun 2017
Posts: 83
Re: For any a and b that satisfy |a – b| = b – a and a > 0, then |a + 3| +  [#permalink]

### Show Tags

13 Oct 2017, 08:41
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| +   [#permalink] 13 Oct 2017, 08:41

Go to page    1   2    Next  [ 22 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

 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®.