how is this A?

We can also solve this by taking some example as there are only four cases possible.
Since, ab > 0
=> a and b have same sign but c could have any sign.
(Note : c could not be zero because a and be could not be zero and |c| > |a + b|)

So, only four possible cases are
1. a<0, b<0, c<0
2. a<0, b<0, c>0
3. a>0, b>0, c<0
4. a>0, b>0, c>0

[table1b=][/table1b]

a | b | c | Quantity A | Quantity B
-1 | -1 | -3 | 1 | -1
-1 | -1 | 3 | 5 | -1
1 | 1 | -3 | 5 | -1
1 | 1 | 3 | 1 | -1

So, in all the cases Quantity A > Quantity B.

How the explanations above are explaining ab<0 and in the original question ab>0?

Explanation:
1- Since ab>0, a and b have the same sign.
2- Since |a+b| = |a|+|b|, then the first two terms in a and b are equal.
3- For C :
we will use choosing numbers strategy :
let a=-1, b:-2, c=-5 ( satisfying the rule )

then :
-1-2-5= -8, |-8| will equal 8
Secondly,
1+2-5= -2
Answer A.

But the problem with choosing # that we cannot grant our result is right. Therefore, we will need to choose other number. So, C= -4
A= 3-4= -1
B= -3-4=-7---> |7|=7

Then, we can now be sure more than our choice is the right choice. Answer: B

My bad, I have just mistyped ab>0 as ab<0. Rest of the explanations are correct as they are.

Sorry but the OA is A. How can be B ??

ab>0 is possible when

1. a and b are positive
2. a and b are negative

1. a and b are positive

let us assume values a=2 b=5

|a + b| = 7

|c| > 7

we will take 2 values of c
A. c = 10
B. c = -10

A. a=2 b=5 c=10

Quantity A

|a + b - c| = 3

Quantity B

|a| + |b| - |c| = -3

Quantity A > Quantity B

B. a=2 b=5 c = -10

Quantity A

|a + b - c| = 17

Quantity B

|a| + |b| - |c| = -3

Quantity A > Quantity B

Similarly

2. a and b are negative

let us assume values a=-2 b=-5

|a + b| = 7

|c| > 7

we will take 2 values of c
A. c = 10
B. c = -10

A. a= -2 b= -5 c=10

Quantity A

|a + b - c| = 17

Quantity B

|a| + |b| - |c| = -3

Quantity A > Quantity B

B. a= -2 b= -5 c = -10

Quantity A

|a + b - c| = 3

Quantity B

|a| + |b| - |c| = -3

Quantity A > Quantity B

Its a simple problem. Take smart numbers.
Given:
ab>0
|c|>|a+b|

First statement means a and b both should have same sign only then we will get positive answer greater than zero so assume a=2 and b=3 (I have taken positive values)
Next |c|>|a+b|
if a = 2 and b = 3 so a+b=5 and we can take c=6

Quantity A: |a+b−c|
|2+3-6|=|5-6|=|-1|=1 (outside mod the value will be positive)

Quantity B: |a|+|b|−|c|
|2| + |3| - |6|
2+3-6
=5-6
=-1

So Quantity A is greater as it is +1

See more on our math book absolute value theory

https://gre.myprepclub.com/forum/gre-quant ... tml#p52039

Ask if something is still unclear to you

For a = 2, b = 2 and c = 5,
Quantity A = -1, Quantity B = -1
A=B

For a = 2, b = 2 and c = -5,
Quantity A = 9, Quantity B = -1
A>B

So, shouldn't the answer be D?

General Information.

We don't know the signs for any of the variables, however, a&b have the same sign because two positive numbers multiplied by each other will give a positive number. The exact same situation happens for 2 negative numbers, their multiplication results in a positive number.


Quantity A Evaluation

a+b-c will give a non-zero positive or negative value since |c| > |a+b|. The sign of a and b do not matter because as long as the absolute value of their sum is not equal to c, the result will be a non-zero value irrespective of the sign.

The absolute value of the result is then positive.

Hence Quantity A, |a+b-c| is positive


Quantity B Evaluation

Because a and b have the same sign and |c| > |a+b|, hence |c| > |a| + |b|

The sum each of the absolute values of a and b will still be less than the absolute value of c,
so |a|+|b|-|c| will give a non-zero negative value.

Hence Quantity B, |a+b-c| is negative


Answer hence is A and Quantity A is bigger.

For a = 2, b = 2 and c = 5, quantity A should be 1, because of the absolute sign.

ab>0 so both a and b are either > 0 or < 0
so |a| + |b| = |a+b|
since |c| > |a+b| then |a| + |b| - |c| is < 0
|a+b-c| always >= 0 then its A