Bunuel wrote:

If \(abc ≠ 0\), is \(abc > 0\) ?

(1) \(|a –b | = |a| - |b|\)

(2) \(|b + c| = |b| + |c|\)

Let's take 2 & -2 and plug them across the statements (or choose any 2 values). I choose a number and its apposite to make calculations easy.

(1) \(|a –b | = |a| - |b|\)

(a,b) = (2, 2)

\(|2 –2 | = |2| - |2|\)= 0

(a,b) = (-2, -2)

\(|-2 +2 | = |2| - |2|\)= 0

(a,b) = (-2, 2) and (2,- 2) are invalid as RHS does not equal LHS

Conclusion is a & b must have same sign but c can be any number with any signInsufficient

(2) \(|b + c| = |b| + |c|\)

(b,c) = (2, 2)

\(|2 + 2| = |2| + |2|\)= 4

(b,c) = (-2,-2)

\(|-2 -2| = |-2| + |-2|\)= 4

(b,c) = (-2, 2) and (2,- 2) are invalid as RHS does not equal LHS

Conclusion is b & c must have same sign but a can be any number with any signInsufficient

Combining 1 & 2

Case I: 2, 2 , 2.................Answer is Yes

Case II: -2, -2,-2..............Answer is NO

Insufficient

Answer: E