Postal wrote:
if zy<xy<0 is Ix-zI+IxI=IzI
1) Z<x
2) Y>0
Instead of plugging in numbers, use the conceptual understanding of mods and inequalities here.
Important points to remember:
1. If ab is negative, either a or b but not both should be negative.
2. If a is positive, |a| = a; if a is negative, |a| = -a
3. If |a| > |b| but a < b, a must be negative.
Given: zy<xy<0
This means zy and xy are both negative. This is possible only if [highlight]either y is negative (and z and x are positive)[/highlight][highlight][/highlight] OR [highlight]both z and x are negative but y is positive[/highlight]. Also, since zy is 'more negative' than xy, the absolute value of z is higher than that of x i.e. [highlight]|z| > |x|[/highlight]
Question: Is |x-z| = |z| - |x|?
Let's look at stmnt 2 first. y > 0
This means that both x and z are negative. We know that if z is negative, |z| = -z. Therefore, the question now is: [highlight]Is |x-z| = x - z[/highlight]?
Again, we know that |x-z| is equal to x-z if x-z is positive. So, is x-z positive? We know that x and z are both negative and z is more negative than x (e.g. x = -2 and z = -3). Therefore, x-z will be positive.
So the answer to Is |x-z| = x - z? is 'YES'
Sufficient.
Stmnt 1: z < x
We know from above that |z| > |x|. If still z < x, then z must be negative (which implies that both x and z must be negative). This is exactly same as the situation above in stmnt 2 hence this statement will be sufficient alone.
Answer D.
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Karishma
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