IMO :A

if b>a, |a|-|b| will always be negative while |a-b| always positive, so relations is established

1, b>a, b=5, a=3, |a-b| = |3-5| = 2, |a|-|b| = 3-5 = -2, so |a-b| > |a-b|
b=5, a=-3, |a-b| = |5-(-3)| = 8, |a|-|b| = -8

Answer seems to C

If you combine two inequalities b>a>0 then b will be always greater than a and it will +ve. Now |a|-|b| will be alway smaller than |a-b| because the |a|-|b| will be always -Ve and since this is YES/NO the answer will be alway "no"

Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me

Anurag@Gurome wrote:

koby_gen wrote:Is |a| - |b| >= | a - b | ?

(1) b > a
(2) a > 0

Statement 1: b > a
Say, a = 0 and b = 3 => |a| - |b| = -3 < |a - b| = 3
Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3

Not sufficient

Statement 2: a > 0
Say, a = 1 and b = 2 => |a| - |b| = -1 < |a - b| = 1
Say, a = 1 and b = 0 => |a| - |b| = 1 = |a - b| = 1

Not sufficient

1 & 2 Together: b > a > 0
Hence, |a| - |b| < 0
But, |a - b| > 0

Hence, |a| - |b| < |a - b|

Sufficient

The correct answer is C.

1. b>a

values i picked

a b
2 3
-3 -2

i found that for any vlaue of a and b when b>a, | a| - |b| turned out to be <= |a-b|

hence sufficent

similarly for

2. a> 0

a b
2 1
2 4
2 -1
2 0

again for all values of a,b, |a| - |b| turned out to be <= |a-b|

heance sufficent. I would have picked D. Either statement is sufficent. Please correct my thought process if it is wrong. Thanks

madhan_dc

for st 2, if a=2 b=-1, then | a| - |b| < |a-b|
but if a=2 b=1 then | a| - |b| = |a-b|
hence not suff.

madhan_dc wrote:1. b>a

values i picked

a b
2 3
-3 -2

i found that for any vlaue of a and b when b>a, | a| - |b| turned out to be <= |a-b|

hence sufficent

similarly for

2. a> 0

a b
2 1
2 4
2 -1
2 0

again for all values of a,b, |a| - |b| turned out to be <= |a-b|

heance sufficent. I would have picked D. Either statement is sufficent. Please correct my thought process if it is wrong. Thanks

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me

Ok try a=3 & b= 4....

Does ur stmt hold true??

ofcourse,
even wth a=3,b=4
|a| - |b| < |a - b|
nd this would hold true for all values of b>a.

gmatmachoman wrote:

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me

Ok try a=3 & b= 4....

Does ur stmt hold true??

@Anurag, How can 1 be > 3? You have stated so and I quote.
Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3?

According to me, Each statement alone is sufficient.
Logic: |a-b| is always a +ve number
Let us analyze |a| - |b|, based on the two statements.
1) b> a. In this case |a|-|b| can become negative when both a & b are +ve. Hence this is sufficient to say that
|a|-|b| is not always >= |a-b|.
2) a>0. In this case |a|-|b| can be negative when b>a ie when a & b are +ve. Hence this statement is sufficient to say that |a|-|b| is not always >= |a-b|.
So the answer is Each Statements Alone is Sufficient.
Correct me if i'm wrong.

Thanks,
FK

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me

Anurag@Gurome wrote:

koby_gen wrote:Is |a| - |b| >= | a - b | ?

(1) b > a
(2) a > 0

Statement 1: b > a
Say, a = 0 and b = 3 => |a| - |b| = -3 < |a - b| = 3
Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3

Not sufficient

Statement 2: a > 0
Say, a = 1 and b = 2 => |a| - |b| = -1 < |a - b| = 1
Say, a = 1 and b = 0 => |a| - |b| = 1 = |a - b| = 1

Not sufficient

1 & 2 Together: b > a > 0
Hence, |a| - |b| < 0
But, |a - b| > 0

Hence, |a| - |b| < |a - b|

Sufficient

The correct answer is C.

IMO A is sufficient.

Please post OA

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me



[

Ok, u took b>a. but what if a<b<0??

a= -5 b= -4

then LHS = 1 & RHS = 1. So the prompt becomes YES

case 2 : a= 3 b = 5

LHS : -2 & RHS will be a positive number 2. Now it says NO.

So st1 is not sufficient.

Is |a| - |b| >= | a - b | ?

(1) b > a

(2) a > 0

okay let me analyse all the cases possible here.

st 1 b>a
for b>a>0, LHS is always < RHS. b=5, a=3. LHS=-2, RHS=2.
for a<b<0. a=-5,b=-2. LHS=3, RHS=3. LHS=RHS. for all negative values, LHS=RHS.
so form 1 we get, |a| - |b| < =| a - b |
the question asks whether |a| - |b| >= | a - b |,
im confused here as to wht shud the answer choice be?
shudnt it be A?

gmatmachoman wrote:

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me



[

Ok, u took b>a. but what if a<b<0??

a= -5 b= -4

then LHS = 1 & RHS = 1. So the prompt becomes YES

case 2 : a= 3 b = 5

LHS : -2 & RHS will be a positive number 2. Now it says NO.

So st1 is not sufficient.

@rohu
st(1) b>a means that b-a>0 OR a-b<0. So we need to validate |a|-|b| >=0 OR |a|>=|b| OR a^2>b^2 Is this true? No, because it says in st(1) that b>a Not Sufficient // do you agree?
st(2) a>0. So we need to validate a - |b| >= |a-b| OR a>= |a-b| + |b|
case 1) (a-b)>0, a>b and b>0 .... a>a-b+b, 0>0 this is invalid determination;
case 2) (a-b)=0, a=b and since a>0 then b>0 .... a>0+b, a>b this contradicts to a=b, hence invalid determination;
case 3) (a-b)<0, a<b and since a>0 then b>0 .... a>b-a+b, a>b this contradicts to a<b, hence invalid determination;
statement (2) proved Not Sufficient;
Combined st(1&2): a>0 and b>a, hence b>0 --> a-b >=|a-b| ? if a-b>0 then a>b canceled, if a-b<0 then a<b valid, if a-b=0 then a=b canceled --> we keep only one condition for |a-b| which a-b<0 Sufficient

answer C

rohu27 wrote:Is |a| - |b| >= | a - b | ?

(1) b > a

(2) a > 0

okay let me analyse all the cases possible here.

st 1 b>a
for b>a>0, LHS is always < RHS. b=5, a=3. LHS=-2, RHS=2.
for a<b<0. a=-5,b=-2. LHS=3, RHS=3. LHS=RHS. for all negative values, LHS=RHS.
so form 1 we get, |a| - |b| < =| a - b |
the question asks whether |a| - |b| >= | a - b |,
im confused here as to wht shud the answer choice be?
shudnt it be A?

gmatmachoman wrote:

rohu27 wrote:Say, a = -2 and b = 1 => |a| - |b| = 1 > |a - b| = 3
even in the above |a| - |b| < |a - b|
for any value of b>a it holds true.
so the answer shud be A rite?
im i missing sumthgn here, inequalities combined wth absolute values in DS is a deadly combo for me



[

Ok, u took b>a. but what if a<b<0??

a= -5 b= -4

then LHS = 1 & RHS = 1. So the prompt becomes YES

case 2 : a= 3 b = 5

LHS : -2 & RHS will be a positive number 2. Now it says NO.

So st1 is not sufficient.

aah...get it now.
thanks gmatmachoman for pointing it out
and Nightreader for a diff approach for such prblems.