Bunuel wrote:
There are two ways to go about it.
First way is by opening and solving the modulus.(1) \(|a - b| < |b|\)
\(1]\) \(a - b < b\) = \(a < 2b\) => This means that either
(a is negative and b is positive) or (both a and b are positive) - plugin any value and check.
\(2]\) \(a - b < -b\) = \(a < 0\) => a is positive
\(3]\) \(-a + b < b\) = \(-a < 0\) = \(a>0\) => a is negative
\(4]\) \(-a + b < -b\) = \(a > 2b\) => This means that either (a is positive and b is negative) or
(both a and b are negative) - plugin any value and check.
From here we know that a and b can anything +ve or -ve. Statement 1 is insufficient.
(2) \(a < 0\)
We don't know anything about b.
(1) + (2) tells us that it's either one of the above highlighted cases. Again, insufficient.
Second way to go about it is by squaring both sides.(1) \(|a - b| < |b|\)
\((a - b)^2 < b^2\) = \(a^2 + b^2 - 2ab < b^2\) = \(a^2 - 2ab < 0\) = \(a(a - 2b) < 0\)
Now, let's think about this
If \(a(a - 2b) < 0\), then either \(a<0\) or \(a-2b<0\)
a<0 doesn't tell anything about b
\(a<2b\) tells us that either
a is negative and b is positive or both a and b are positive or both can be negative as well (Eg: a = -10, b = -4). Only then \(a<2b\) will hold true. It is very easy to miss out on the fact that both a and b can be negative when following this approach.
Hence, Insufficient.
(2) \(a < 0\)
Doesn't tell us anything about b. Insufficient.
(1) and (2)
Combining them still doesn't tell us anything about b - it can be both +ve and -ve as in the highlighted portion above.
Therefore, E should be the correct answer choice. _________________
Vaibhav
Sky is the limit. 800 is the limit.
~GMAC