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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.

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. -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

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, 04:39

1

This post received KUDOS

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

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

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

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.

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

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.

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.

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

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).

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).

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.

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....