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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


Is c > 2 if a + b + c > 0 ?

(1) c > a + b + 2

(2) a + b + 2 < 0


Transforming the original condition and the question, we have c>-a-b from c>2? and thus we need to know the relation between -a-b and 2. From 2), we find that a+b<-2 and a+b+c>0, therefore -c<a+b<-2, -c<-2, c>2 thus it is sufficient, and B is the answer.
In case of 1), if c = 5 then a=b=1 and the answer is yes, but if c =2 then a=b=-0.5 and the answer is no. Thus it is not sufficient.
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In statement (1) and (2) we have a+b common and Question also have a+b
To simplify, assume a+b=x

So,the Question becomes Is c>2 if x+c>0?

Statement (1)c>x+2,Sum of x+c>0 and c>x+2 (we can sum inequalities if their signs are in the same direction):

x+c+c>0+x+2

or,c>1

Not Sufficient

Statement (2) x+2<0,Subtract x+2<0 from x+c>0 (we can subtract inequalities if their signs are in the opposite directions):

x+c-x-2>0

or, c>2


Sufficient

So the Correct Answer is B
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Solution:

a+b+c > 0 ==> c > -(a+b) -->(1)

Statement1 : c > a + b + 2 --> (2)
Add (1) and (2), then 2c > 2 --> c > 1
Therefore, insufficent.

Statement2 : a + b + 2 < 0 ==> 2 < -(a+b) ==> -2 > a+b -->(3)
add (1) and (3),then c-2 > 0 ==> c > 2
Sufficient.

So, Option B
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