Is ab 0? (1) |a + b| ≠ |a| + |b| (2) a

Question

Is ab > 0?

(1) |a + b| ≠ |a| + |b|
(2) a < b

deabas · Answer

Is ab > 0?

(1) |a + b| ≠ |a| + |b|
If try different values (+ve and -ve) for both a and b, then this statement is true if and only if a and b are different in signs
---------> ab<0
Statement 1 is sufficient (cross out B, C, and E)

(2) a < b
It depends on the values of both a and b (+ve and/or -ve)
Statement 2 is not sufficient (Cross out B)

The final answer is "A"

ishan92 · Answer

As per the property, |x+y|≤|x|+|y|.
Since in (1), it is given that they are not equal, so they must be less than,ie, |a+b|<|a|+|b|.
This is can only happen when a and b are of opposite signs.
Hence ab<0.
(1) is sufficient.

Clearly (2) is insufficient on its own.

ShivamoggaGaganhs · Answer

Here,

a and b both must be pos or neg

mod(a+b)!= mod(a)+ mod(b)

possible only when a and b are of opposite signs.

since they are of opposite sign, 
ab<0 ----sufficient

2)N.S.

blanchif · Answer

Hello

Could someone explain a little more this question...I do not understand why you come up with this answer.

Thanks for the help.

CEdward · Answer

Is ab > 0?

(1) |a + b| ≠ |a| + |b|
(2) a < b

ab > 0 would imply that a and b each have the same sign. So if we can show that a and b have opposite signs, then that would address the question.

(1) |a + b| ≠ |a| + |b| Implies that a and b have opposite signs (i.e. ab = -ve)
Sufficient

(2) a < b
Both and and b could be positive or they could have opposite signs. Two different possibilities
Insufficient

Is ab > 0? (1) |a + b| ≠ |a| + |b| (2) a < b

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