bharat2384 wrote:
If zy<xy<0 is |x-z|+|x|=|z|
1.z<x
2.y>0
This question was discussed before here is my post from there:
This is not a good question, as neither of statement is needed to answer the question, stem is enough to do so.
If \(zy<xy<0\) is \(|x-z|+|x| = |z|\)
Look at the inequality \(zy<xy<0\):
We can have two cases:
A. If \(y<0\) --> when reducing we should flip signs and we'll get: \(z>x>0\).
In this case: as \(z>x\) --> \(|x-z|=-x+z\); as \(x>0\) and \(z>0\) --> \(|x|=x\) and \(|z|=z\).
Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(-x+z+x=z\) --> \(0=0\), which is true.
And:
B. If \(y>0\) --> when reducing we'll get: \(z<x<0\).
In this case: as \(z<x\) --> \(|x-z|=x-z\); as \(x<0\) and \(z<0\) --> \(|x|=-x\) and \(|z|=-z\).
Hence in this case \(|x-z|+|x|=|z|\) will expand as follows: \(x-z-x=-z\) --> \(0=0\), which is true.
So knowing that \(zy<xy<0\) is true, we can conclude that \(|x-z|+|x| = |z|\) will also be true. Answer should be D even not considering the statements themselves.
As for the statements:Statement (1) says that \(z<x\), hence we have case B.
Statement (2) says that \(y>0\), again we have case B.
Hope it helps.
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