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29 Jul 2015, 03:28
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Is |x-z| + |x| = |z|?
(1) zy < xy < 0
(2) y > 0
(1) zy < xy < 0
(2) y > 0
Is there a faster way other than trying out different positive / negative cases for 1?
29 Jul 2015, 03:53
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noTh1ng
Is |x-z| + |x| = |z|?
(1) zy < xy < 0
(2) y > 0
(1) zy < xy < 0
(2) y > 0
Is there a faster way other than trying out different positive / negative cases for 1?
Is |x-z| + |x| = |z|?
(1) zy < xy < 0. Consider two cases:
(i) If y > 0, then we'd get: z < x < 0.
z < x means that x - z > 0, thus |x - z| = x - z;
x < 0 means that |x| = -x;
z < 0 means that |z| = -z.
Therefore, LHS = |x - z| + |x| = (x - z) - x = -z and RHS = |z| = -z --> -z = -z.
(ii) If y < 0, then we'd get: z > x > 0 (flip the sign when reducing by negative value).
z > x means that x - z < 0, thus |x - z| = -(x - z);
x > 0 means that |x| = x;
z > 0 means that |z| = z.
Therefore, LHS = |x - z| + |x| = -(x - z) + x = z and RHS = |z| = z --> z = z.
Since in both case we got an YES answer to the question, then this statement is sufficient.
(2) y > 0. Clearly insufficient.
Answer: A.
Hope it's clear.
29 Jul 2015, 07:33
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noTh1ng
Is |x-z| + |x| = |z|?
(1) zy < xy < 0
(2) y > 0
(1) zy < xy < 0
(2) y > 0
Is there a faster way other than trying out different positive / negative cases for 1?
|x-z| + |x| = |z| will always be true when either z < x < 0 or z > x > 0 or when x=0
Statement 1: zy < xy < 0
This will be true only when y is positive and z < x < 0
or This will be true only when y is Negative and z > x > 0
Hence,
SUFFICIENT
Statement 2: y > 0
Nothing Can be predicted about x and z hence
NOT SUFFICIENT
Answer: option A
