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(1) |a| > |b| means that the distance of 'a' from zero is greater than the distance of 'b' from zero, on the number line. But both could be positive (in which case a > b and a+b>0), both could be negative (in which case a < b and a+b<0) or a positive, b negative (in which case a>b and a+b>0) or a negative, b positive (in which case a<b and a+b<0).
So Insufficient.

(2) a < b
Again this doesnt tell us whether a+b is positive or negative. Insufficient.

Combining the two statements, if a<b and |a|>|b|, it could happen in two cases:
I) either both are negative but a is more away from zero than b
II) or a is negative, b is positive but distance of a from zero is more than distance of b from zero

In either of the two cases, sum will be negative. Or a+b<0. Sufficient.

Hence C answer
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chetan2u
Is a + b > 0?

(1) |a| > |b|
(2) a < b

Awesome question! seriously Chetan this is easy but tricky
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chetan2u
Is a+b>0?
(1) |a|>|b|
(2) a<b

From Stmnt 1 (a+b)(a-b)>0. So insufficient.
From Stmn 2 (a-b) < 0. Insufficient.

Combining 1 & 2...

a+b<0.

Please let me know if anything wrong with my approach.

Really love this approach; can anyone confirm if this method is acceptable in the GMAT?
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Is a+b>0?

(1) |a|>|b|

a = 3, b =2 YES
a = -3, b = 2 NO

Insufficient

(2) a<b

Clearly insufficient.

Combo pack:
a must have a greater absolute value relative to b, but still be less in (non-absolute) value. In other words, a + b is never > 0.
Sufficient.
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chetan2u
Is a + b > 0?

(1) |a| > |b|
(2) a < b

Self made - tricky


Number line
(1) |a| > |b|
The distance of a from 0 is more than the distance of b from from 0.
Both a and b could be positive...Yes
Both could be negative....No

(2) a < b
Both a and b could be positive...Yes
Both could be negative....No

Combined
a<b means that a should be closer to 0, if a and b are positive. But we know that a is farther from 0 from statement I. Hence a is surely negative.
Now, a is negative and farther from 0, whatever be b, the sum a+b will be less than 0 as |a| has a higher value.

Algebraic
(1) |a| > |b|
As both sides are positive, we can square the two sides
\(|a|^2 > |b|^2...............a^2>b^2....a^2-b^2>0.......(a+b)(a-b)>0\)
Nothing more as a+b will depend on a-b.

(2) a < b
\(a-b<0\)
Nothing about a+b

Combined
\((a+b)(a-b)>0\), so a+b and a-b will have same sign.
a-b<0, so a+b is also less than 0.
Answer is NO

C
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