iamseer wrote:If a and b are integers, and |a| > |b|, is a · |b| < (a - b)?

(1) a < 0

(2) ab >= 0

a · |b| =>>>> is it product?

Confirm?

Yes.

Once again the question is:

If a and b are integers, and |a| > |b|, is a*|b| < (a - b)?

(1) a < 0

(2) a*b >= 0

Statement 1:
|a| > |b| and a < 0
Assume a = -5, b = +4 or -4

a*|b| < (a - b)
-5 * 4 < -5 - (-4 or +4)
-20 < -1 or -9 TRUE

Assume a = -1/2, b = +1/3 or -1/3
a*|b| < (a - b)
-1/2 * 1/3 < -1/2 - (-1/3 or +1/3)
-0.166 < -0.834 or -0.166 NOT TRUE

Not Sufficient

Similarly try for others

akhp77 wrote:Statement 1:
|a| > |b| and a < 0
Assume a = -5, b = +4 or -4

a*|b| < (a - b)
-5 * 4 < -5 - (-4 or +4)
-20 < -1 or -9 TRUE

Assume a = -1/2, b = +1/3 or -1/3
a*|b| < (a - b)
-1/2 * 1/3 < -1/2 - (-1/3 or +1/3)
-0.166 < -0.834 or -0.166 NOT TRUE

Not Sufficient

Similarly try for others

Why are we assuming A and B as fractions ? The question requires them to be integers right ? OR am I missing some thing important here ?

iamseer wrote:If a and b are integers, and |a| > |b|, is a · |b| < (a - b)?

(1) a < 0

(2) ab >= 0

Source: MGMAT CAT
OA : E

What is the best way to tackle such questions? Is there some technique to handle these kind of absolute value questions better?

Thanks.

The best way to solve these kind of question is squaring both sides .

Lets look at the question .

We have |a| > |b| --- squaring both sides

a^2 > b ^2 ----> a^2 - b^2 > 0 .

Now we need to know is a · |b| < (a - b)?

Thats a yes / no question . Again square both sides

a^2 b^2 < a^2 - 2ab - b^2

a^2 b^2+ 2ab < a^2 - b^2

so the question becomes is ab(ab+2) < a^2 - b^2 ?

(1) a < 0

Nothing is said about ' b' so insufficient .


(2) ab >= 0

Now if ab = 0 we get

a^2 - b^2 > 0 that is same as what is given in question so nothing new .

But if we do consider ab > 0 . We will get an answer .

So again we can not say for sure .

Insufficient .

Lets combine 1 and 2 .

we know a = negative so b is also negative as ab >0 from 2 .

But it dose not matter if a and b are positive or negative it will lead to the same situation as in statement 2 . Where ab could be equal to zero .

hence this too is not sufficient .

Ans E.

rockeyb - Yes you are right. a and b are integer

Got this problem wrong. But found an easy approach after the test. Wanted to share with all.

Need to check whether a*|b|<a-b
-> a*|b| -a +b<0
a*(|b|-1)+b <0

stmt1 says a<0.
B can be 0 or <0 or >0
Assume B to be 0,1,-1

Substitute and check if it works in the eqtn a*(|b|-1)+b <0

when b=0,
a<0, hence -a>0
-a+0 >0
-a>0... NO

when b=-1
-1<0..YES

No need to check b=1. one YES and one NO.
hence INSUFF

(2) ab >= 0
a and b can both be negative or a is negative and b=0
These 2 cases We have already covered in stmt1.
hence INSUFF

Combining:
We still have the same failing case.
hence INSUFF

pick E.

iamseer wrote:If a and b are integers, and |a| > |b|, is a · |b| < (a - b)?

(1) a < 0

(2) ab >= 0

Source: MGMAT CAT
OA : E

What is the best way to tackle such questions? Is there some technique to handle these kind of absolute value questions better?

Thanks.

Plugging in values works well if you choose wisely. To be efficient, search for values that satisfy both statements, recognizing how the problem is restricted -- and how it isn't. Here's what would go through my head as I read through the problem:

a and b are integers. Ok, no decimals.
|a| > |b|. Ok, a has to be farther from 0 than b.
Looking at statement 1: a<0. Ok, only negative values for a.
Looking ahead to statement 2: ab>=0. Ok, to satisfy both statements, b must be negative or 0.

So let's try a = -2, b = -1, and b=0. These values satisfy all the conditions.

If a = -2 and b = -1:
Is a · |b| < (a - b)?
-2 * |-1| < (-2 - (-1))
-2 < -1 Yes.

If a = -2 and b=0:
Is a · |b| < (a - b)?
-2 · |0| < (-2 - 0)
0 < -2 No.

Since the answer can be both yes and no, and since we've chosen values that satisfy both statements, the correct answer is E.

By recognizing how the problem is restricted -- and how it isn't -- and by choosing values that satisfy both statements, we make the process much more efficient.