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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