If A and B are non-zero numbers such that AB >0, is |A - B| > |A| - |B

If A and B are non-zero numbers such that AB>0, is |A−B|>|A|−|B|?

A(+)/B(+) OR A(-)/B(-) both have same sign.
Three things to take care while solving:
1. Sign of A and B.
2. Whether A>B OR A<B
3. Whether A and B are fractions OR one of them is a fraction

(1) \(\frac{1}{A}<\frac{1}{B}\)
Multiplying both sides by AB as AB>0
\(\frac{AB}{A}<\frac{AB}{B}\)
B < A
But nothing about the signs and fractions
Case I: A = 6, B = 2; |6−2|>|6|−|2| i.e. 4 > 4 NO
Case II: A = -2, B = -6; |-2−(-6)|>|2|−|-6| i.e. 4 > -4 YES

INSUFFICIENT.

(2) 2A+B<0
Hence both A and B are negative. But whether A>B OR A<B and fractions ?
Case I: A = -6, B = -2; |-6−(-2)|>|-6|−|-2| i.e. 4 > 4 NO
Case II: A = -2, B = -6; |-2−(-6)|>|2|−|-6| i.e. 4 > -4 YES

INSUFFICIENT.

Together 1 and 2
All the three conditions are known now. Sufficient then. Just to verify.
Case II of statements applicable still. YES case.
Case III: \(A = -\frac{1}{3}, B = -\frac{1}{2}; |-\frac{1}{3}−(-\frac{1}{2})|>|-\frac{1}{3}|−|-\frac{1}{2}| i.e. \frac{1}{6} > -\frac{1}{6}\) YES

SUFFICIENT.

Answer C.
PS: Case III would have been only necessary had the signs differed with some other conditions. Thus, it's unnecessary in this question to check fractions.

AB > 0 --> A and B have same sign

Q. is |A - B| > |A| - |B|?

NOTE:
If A>B and both are +ve, then |A - B| = |A| - |B|
If A>B and both are -ve, then |A - B| > |A| - |B|

If A<B and both are +ve, then |A - B| > |A| - |B|
If A<B and both are -ve, then |A - B| = |A| - |B|

From above observation, one have to know whether A>B and whether both A, B are +ve or -ve

(1) 1/A < 1/B --> A > B
But we don't know the sign of A and B (both are +ve or -ve)
NOT SUFFICIENT

(2) 2A + B < 0 --> -B > 2A or B < -2A
- A and B cannot be +ve, so A and B must be -ve
- B < -2A, Bu we are not sure whether A>B.

(1)+(2)
A > B ; A, B are both -ve
So, as discussed above , |A - B| > |A| - |B|

FINAL ANSWER IS (C)

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If A and B are non-zero numbers such that AB>0, is |A−B|>|A|−|B|?

(1) 1/A<1/B

(2) 2A+B<0

Given: A,B \(\ne0\), A,B can be (+,+) or (-,-).

statement 1:
1/A < 1/B
A \(\ne\) B
possible cases:

(+,+): A > B then |A−B|>|A|−|B| is false; consider (4,2) or (0.5,0.3)
(-,-): |B| > |A| then |A−B|>|A|−|B| is true; consider (-2,-4) or (-0.3,-0.5)

not sufficient

statement 2:
2A + B < 0
A,B \(\ne\)(+,+)
possible cases

(-,-): |A| > |B| then |A−B|>|A|−|B| is false; consider (-4,-2)
(-,-): |B| > |A| then |A−B|>|A|−|B| is true; consider (-2,-4)

not sufficient

combining both statements,
only possible case : (-,-) & |B| > |A|.
sufficient
Ans: C

ab>0 = same signs
|a-b|>|a|-|b| when a<b
everything else, condition is false

(1) sufic

a b 1/a < 1/b
3 2 1/3 < 1/2 a>b
-3 -2 -1/3 < -1/2 invalid
-1/3 -1/2 -3 < -2 a>b
1/3 1/2 3 < 2 invalid

(2) insufic

2a+b<0
a,b=-1,-1: -2-1<0 a=b
a,b=-2,-1: -4-1<0 a<b
a,b=-1,-2: -2-2<0 a>b

ans (A)

target is |A−B|>|A|−|B|
given AB>0
possible when both ab are +ve or both are -ve
#1
1/A<1/B
B/A<1
means A>B
since given that both (a,b) are + or -ve
case 1 ; both a & b are +ve
a = 2 and b = 1
|A−B|>|A|−|B|
2-1=2-1 ; no
case 2 both a&b are -ve
a=-1 and b = -2
l-1+2l>l-1l-l-2l
1>-1 ; yes
insufficient
#2
2A+B<0
given that both (a,b) are + or -ve
a = b=1
|A−B|>|A|−|B|
no
and both are -ve
a=-1 and b = -2
l-1+2l>l-1l-l-2l
1>-1 ; yes ; insufficient
from 1 &2
only possible case is that both a&b are -ve integers and a>b
OPTION C ; sufficient

If A and B are non-zero numbers such that AB>0, is |A−B|>|A|−|B|?
(1) 1/A<1/B

(2) 2A+B<0

Tough one for me .
Is the ans C ?
Below is my thought process .
If A and B are non-zero numbers such that AB>0, is |A−B|>|A|−|B|?
So both A and B have same sign . Both positive or both negative.
|A−B|>|A|−|B|?
Distance of A to B > distance of A to zero - distance of B to zero ?
This is only possible if |A| < |B| .
So we need to find out |A| < |B| ?

(1) 1/A<1/B
Case 1 :
If A ,B > 0
B < A
|B| < |A| .

Case 2 :
If A , B < 0 .

1/A < 1/B
lets put a value .
-1/2 < -1/3
so A = -2 B = -3
so B < A
but
|B| > |A|..
That's why we cannot conclusively tell that |A| < |B| . SO insufficient .

(2) 2A+B<0
=
A < -B/2 .

Lets take B as 1 , A < -1/2 (Not possible since A B both same sign .)
lets take B = -2 . A < -1 . A can be -1.9 ,-2 or -3 .
So for different value of A we can get different relation of |A| < |B| or |A| = |B| or |A| > |B|
So not sufficient .

Now combing 1 and 2 .
A , B < 0
and |B| > |A|.
so we can conclusively tell that |A−B|>|A|−|B| .
since B = -2 and A = -1.2
0.8 > 1.2 - 2 => 0.8 > -0.8 .
So option C could be my answer .

What an intriguing question! There's a lot going on here. Let's slow down and unpack it bit by bit.

First, we have AB>0. That means A and B have the same sign: either they're both positive, or they're both negative.

Now, what we want to know: is |A−B|>|A|−|B|?. I spent some time trying to simplify this, but the simplification took a long time, and was even harder to explain than it was to do! So, I think the right approach on test day would be to leave this how it is, write it down on your paper, and get ready to carefully and thoughtfully test cases.

Let's do that now.

Statement 1: 1/A < 1/B.

I'd like to simplify this, but I'm wary, because I know that you usually can't multiply both sides of an inequality by a variable. That's because if you don't know whether the variable is positive or negative, you don't know whether you're supposed to flip the sign or not. But we have something we can actually use here! Because the problem says that AB > 0, we know that AB is positive. So, we can safely multiply both sides of the inequality by AB.

1/A * AB < 1/B * AB
B < A

This statement really says that B < A.

Now let's try some cases where B < A. We're dealing with absolute values, so we should try both positive and negative cases.

Case 1:

B = 1, A = 10. Is |A−B|>|A|−|B|?

|A - B| = |10 - 1| = |9| = 9
|A| - |B| = |10|-|1| = 10 - 1 = 9
Since they're equal, the answer is "no"

Case 2:

B = -10, A = -1. Is |A−B|>|A|−|B|?

|A - B| = |-1 - (-10)| = |9| = 9
|A| - |B| = |-1| - |-10| = 1 - 10 = -9

The left side is bigger, so the answer is "yes."

We got both types of answers, so this statement is insufficient.

Statement 2: 2A+B<0

We know, from the original question, that A and B are either both positive or both negative. They definitely can't both be positive anymore, though, given this statement! They must both be negative.

And, if they're both negative... this statement is definitely true! This will be true for any pair of negative values of A and B.

So really, all this is telling us (in the context of this specific problem and the info we already have) is that A and B are both negative. Let's test some cases.

Case 1: First, reuse 'case 2' from the previous statement. That case gave us a "yes" answer.

Case 2: Now, I'm going to try to find a case that works the other way. It seemed to be important whether A is bigger or whether B is bigger, so this time I'm going to try a case where B is bigger than A. Let's say B = -3, and A = -8. Is |A - B| > |A| - |B|?

|A - B| = |-8 - (-3)| = |-5| = 5
|A| - |B| = |-8| - |-3| = 8 - 3 = 5

They're equal, so the answer is "no."

Since we got both a "yes" and a "no," this statement is also insufficient.

Statements 1 and 2 together:

Let's evaluate what we know at this point.

We know from Statement 2 that A and B are both negative.

We also know from statement 1 that A > B.

Is |A - B| > |A| - |B|?

Since A > B, A - B will be a positive number. Therefore, |A - B| = A - B.

Since A and B are both negative, |A| = -A, and |B| = -B.

The question is really asking: Is A - B > -A - (-B)?

Simplify:

Is A - B > B - A?
Is 2A > 2B?
Is A > B?

We already know that the answer to this question is "yes." So, the statements are sufficient together, and the answer to the problem is C.

Nice one!

If A and B are non-zero numbers such that AB>0AB>0, is |A−B|>|A|−|B||A−B|>|A|−|B|?


(1) 1A<1B1A<1B

(2) 2A+B<0
Since, AB > 0, so both of them are of same sign. Another thing is that the left side will be positive. so if A and B are not both negative, the left side will always be greater than the right one.
1) A > B, when A = 2, B = 0.5. Both sides become 1.5. When A = -5, B =-6, 1 > -1. Two different answers. Not sufficient
2) 2A + B < 0. B < -2A. B is negative, so is A. When A = -1, B = -3. Both sides are equal. when A = -1, B = -1, both sides become 0. as, |-A+B|= |B-A| =|A-B| = A-B and |-A| - |-B|= A -B. Sufficient
B is the answer

|A-B|>|A| - |B|
only possible when A > 0 and B < 0 or A < 0 and B > 0

1. 1/A < 1/B
A > B if A > 0 and B > 0
A < B if A < 0 and B < 0

no other information, not sufficient

2. 2A + B > 0
A should be greater than 0, B should be greater than zero, 2 is already greater than 0
Sufficient

Answer - B

Hello nick1816 how did you conclude that this is the question stem?

\(|A - B| > |A| - |B|\)

Since AB>0, the magnitude of |A-B| is equal to magnitude of |A|-|B|.

Hence only way LHS is greater than RHS, if RHS is negative. [Since LHS is always positive}

|A| - |B| < 0

|A| < |B|

If you still have doubt, you can ask.

(1) \(\frac{1}{A} < \frac{1}{B}\)

Same sign therefore we have no issues
A>B

If A,B<0, |B|>|A|
If A,B>0, |B|<|A|

Since both these results we cannot pin down a particular answer
Clearly insufficient

(2) \(2A + B < 0\)

B<-2A
B must be negative.

The magnitude of B is unknown
Clearly insufficient

However when 1 and 2 is combined weget

A,B<0

THerefore |B|>|A|
Clearly sufficient

Therefore iMO C

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