Hi Saurabh,

Please find my reply underlined to your explanation.

Here the question is asking you whether |x-z-y| > |x-z +y| ?

Manipulating the question itself is not a good idea here.

This is much more simple to solve if you just look out at the statements and take stalk of it and apply PLUGGING – IN.

But i selected D. Please let me know where i am wrong

Two possible cases for absolute value

case 1 :  x – z – y > x – z + y . this is true only when x – z –y > 0

After manipulating it becomes y < 0

case 2:  x – z – y > -(x – z + y) . This is wrong. It should read

x-z-y < – (x-z+y)

Solving it you get z > 0. This is true only when x – z – y < 0

After manipulating and changing the sign, x < z

Now, looking at the s(1), i know it answers the question that x < z and it refers to case 2

S(2) says that (x–z–y) is negative. so it clearly refers to case 2, which means x < z. So, it is sufficient isn’t ? – this is wrong.

Please let me know where i am wrong.

Knowing ,

0 < x < z < y will definitely help to answer the question.

Because when z > 0, then x – z – y < 0. It’s the second case. Answer to the question would be YES.

So statement I is sufficient.

Statement II is not sufficient.

It just says x – z – y < 0

But we dont know whether z > o or not.

You can easily disprove this by Plug – in.

Just try the below values.

x = 2 ,  z = 3 and y =4. Here the answer to the question would be YES.

x = 3 , z = -2 and y = 6. Here the answer to the question would be NO.

Two different answers, so not sufficient.

So the answer is A.

Hope this is clear.