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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