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

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

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)

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.

|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

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.

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).

Yes, thanks. It was a typo.

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

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.

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

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

why b-a>0
Please help me to understand

This is because b>a>0; If b = 3, -b = -3 & |-3|= 3

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.

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