DS - absolute value question

[acecoolan]

If a and b are integers, and |a| > |b|, is a · |b| < a – b?

(1) a < 0

(2) ab 0

a MGMAT question - whats the easiest way to solve this?

OA is E

[stubbornp]

The method which i used for thse q's is picking numbers.

acc to stimulus,|a|>|b|

we can assume these values a=3,-3 b=2,-2,0

acc to stmt 1:a<0

it means a=-3,-4 b=-2,2,0...we are getting different answers by putting these values----Insufficient

acc to stmt 2. ab=0

bcoz |a|>|b|....ie b=0--sufficient...

IMO B...if i am not right,then there should be a mistake in stmt 2.

[acecoolan]

Sorry ...just noticed that part of the question was typed wrong.
Statement 2 should read

ab >= 0

[stubbornp]

Sorry ...just noticed that part of the question was typed wrong.
Statement 2 should read

ab >= 0

ok...so now i ll start from stmt 2 directly

ab>=0

one option is either b=0.....

for greater than 0,both a & b should be of same sign...

a=-3,3 b=-2,2

by putting the values,you ll see different results----insufficient


By combing both,again we are getting more than one answers...

So it should be E...hope it helps.

[raju232007]

statement 1:

a<0
a is negative but it tells nothing about b...

Insufficient

statement 2:

ab>=0
so both a and b should have the same sign or any of the two integers can be equal to 0.
In other words a & b are both positive or negative
Let a=3,b=2( Here we should assume a>b as it is given that |a|>|b|)
a · |b| < a – b
6<1............... Ans is no
Let a=-3,b=-2
-6<-1............. Ans is yes

Insufficient

Combining both the statements we get

a<0...so b should also be negative
and |a|>|b|
Let a=-5 b=-2
a · |b| < a – b
-10<-3......Ans is yes
But we have to consider one more case...statement 2 states that ab>=0
so any of the two integers can be equal to zero..(a=0 or b=0)
we know that a<0(from statement 1)..So lets assume b=0
Now let a=-5 & b=0
a · |b| < a – b
0<-5....Ans is no

So the ans should be E...Let me know if you have any doubts...

[maihuna]

Can somebody please solve it algebraically? Ian please?