Hi, Don't worry. You will learn the things and GMAT absolute value tricks sooooon.

The be low is my approcah for any modulus qtn in GMAT.

Remember.
The meaning of |x-y| is "On the number line, the distance between X and +Y"
The meaning of |x+y| is "On the number line, the distance between X and -Y"
The meaning of |x| is "On the number line, the distance between X and 0".

On the # line, Left to 0 are all the -ve #s and right to 0 are all +ve #s

Original Qtn:
Is |a| + |b| > |a + b| ?

menas "is the SUM of the distance between a and 0, and the distance bewtween b and 0 > the distance bewtween a+b and 0"

let us take some cases to simplify the qtn.

case1: if a and b both are RIGHT to 0 on the # line then a+b will also be RIGHT to zero and even more far away from 0 than a or/and b is/are.

Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = 1 and b = 3 then a+b = 4
a is 1 unit away right to 0
b is 3 units away right to
a+b is 4 units away right to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4

-------------0----------b-----a---------a+b---

case2: if a and b both are LEFT to 0 on the # line then a+b will also be LEFT to zero and even more far away from 0 than a and/or b is.

Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = -1 and b = -3 then a+b = -4
a is 1 unit away left to 0
b is 3 units away left to
a+b is 4 units away left to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4

----a+b------a---------b-------0-------------

HENCE,

for |a| + |b| > |a + b| or
|a| + |b| < |a + b| or
|a| + |b| not= |a + b| to be true,
a and b, on the # line, should NOT be on the same side with respect to 0.

So BASICALLY the qtn is ASKING IF a AND b ARE on DEFFERENT SIDES, with respect to 0, on THE # LINE>

stmnt1: a^2 > b^2
from this we can say that the magnitude of a is > the magnitude of b. magnitude means with out sign.
==> it says that |a| > |b|
FROM THIS,
a and b can be on the same side on the # line with respect to 0

-------0------b----------------a , obviosly a is toofar ways from 0 than b is ==> |a| > |b|

OR,

a can be on a different side than b
-----a-------------0---b------ again, a is toofar ways from 0 than b is ==> |a| > |b|

so we have both the cases...hence stmnt 1 is NOT suff.


stmnt2: |a| × b < 0

|a| is always +ve hence b shud be -ve for the product to be -ve.

so all that stmnt 2 says is b is LEFT to 0 on the # line but it does not say whether a is on LEFT or RIGHT to 0. hence NOT suff.

stmnts 1&2 together
stmnt:1 says that a is too far from zero than b is.
stmnt2 says that b is LEFT to 0 on the # line
hence using both too, we can not say if a and b are on different sides to 0 on the #line or on the on the same sides
hence NOT suff.

ANSWER..."E"

Regards,
Murali.
KUDOS?

clean E here.
b < 0 but a can be < > 0
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Murali,

That was a SUPERB explanation..(Thanks+Kudos+...)^n

|a| + |b| > |a + b|

the expression will be true if "a,b have opposite signs e.g. a<0, b>0 etc"

st1) insufficient. a or b could have same or different sign
st2) this proves that b<0. but it is also possible that a<0. so, insufficient.

E is the ans
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