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

Question

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

ENGRTOMBA2018 · Accepted Answer

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.

balamoon · Answer

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

GMATinsight · Answer

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

ENGRTOMBA2018 · Answer

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)

GMATinsight · Answer

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

ashokk138 · Answer

|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

Bunuel · Answer

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).

EMPOWERgmatRichC · Answer

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:  E

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

CyberStein · Answer

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

Nikkz99 · Answer

Can't A and B be same values i.e positive numbers??

Bunuel · Answer

Yes. |a – b| = b – a implies a ≤ b, so a = b is a valid case.

luisdicampo · Answer

Deconstructing the Question

We need to simplify  |a+3| + |-b| + |b-a| + |ab| .

We are given  |a-b| = b-a  and  a>0 .

Since  |a-b| = b-a = -(a-b) , we must have  a-b \le 0 , so  a \le b .

Because  a>0 , it follows that  b \ge a > 0 , so  b>0 .

Step-by-step

Since  a>0 , we have  a+3>0 , so  |a+3| = a+3 .

Since  b>0 ,  |-b| = b .

Since  b \ge a ,  b-a \ge 0 , so  |b-a| = b-a .

Since both  a  and  b  are positive,  |ab| = ab .

Now add the terms:

(a+3) + b + (b-a) + ab

This simplifies to:

a + 3 + b + b - a + ab = ab + 2b + 3

Answer: E

egmat · Answer

Hi Nikkz99,

Great question! Yes, a and b CAN both be positive numbers, and they CAN even be equal. Let me show you exactly what the constraint tells us and how the problem works.

The condition:  |a - b| = b - a means that b - a must be greater than or equal to  0  (because an  absolute value  is always non-negative, so the right side must be too). This gives us  b ≥ a .

Since we're also told  a > 0 , we know:  b ≥ a > 0  — so YES, both are positive.

Now let's simplify each piece using b ≥ a > 0:

• |a +  3 | = a +  3  (since a >  0 , so a +  3  is positive)
• |-b| = b (since b >  0 )
• |b - a| = b - a (since b ≥ a, so b - a is non-negative)
• |ab| = ab (since both positive, ab is positive)

Adding them up:  (a +  3 ) + b + (b - a) + ab

Note that  the a and -a cancel:  =  3  +  2 b + ab = ab +  2 b +  3

Answer: E

Verification:  You can verify with a = b =  1 :
| 1  +  3 | + |- 1 | + | 1  -  1 | + | 1  ×  1 | =  4  +  1  +  0  +  1  =  6 
E gives:  1  +  2  +  3  =  6  ✓

Key Insight:  |x| = -x doesn't mean the result is negative — it means x itself was negative, so the negative sign flips it positive. Here, |a - b| = b - a tells us a - b ≤  0 , meaning b ≥ a.

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

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