|a| + |b| > |a + b| ? ?
b) |a|* b <0. as |a|>=0 always, b<0..
b=-1,a=-2 -> lhs=rhs.
b=-1, a=2 -> lhs>ths Insufficient

a) a^2>b^2 above 2 cases still apply. Insufficient..
a&b) Insufficient
IMO E

(2) - clearly states b is negative.

if a= 1 and b = -3 then we have |1| + |-2| > | 1-2| => 3 = 1.. not true.
if a = 3 and b = -1 then we have |3| + |-1| > |3 -1| => 4 > 2 not true...

From above is it not possible (2) alone is sufficient...

@vidhya
yes statement 2 does say that b is negative. Which means |a| is +ve. If |a| is +ve we still have no indication whether a is +ve or -ve. Because |a| is gonna be +ve irrespective of the a's sign. In the values that you've plugged in you've only put in +ve values for a and -ve values for b. Try putting in negative for both a and b and the inequality will be flipped. Hence E

E for me as well

bcos even if b is <0 a also can < 0
so both side will add up

In the initial statement:

Is |a| + |b| > |a + b| ?

If a > 0 and b > 0, then
|a| = a
|b| = b
|a + b| = a + b
So |a| + |b| = |a + b|

If a < 0 and b < 0, then
|a| = -a
|b| = -b
|a + b| = -a - b
So |a| + |b| = |a + b|

Only if one is positive and one is negative will |a| + |b| > |a + b|, because when one is positive and one is negative |a + b| will be closer to 0 than the individual |a| + |b|.

So a translation of this question is:
Do a and b have different signs (in which case the inequality is true), or the same sign (in which case they are equal and it is false)?

With this in mind, the statements tell us that b < 0 but nothing about the sign of a.