Bunuel
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