ds4

IMO C

We just have to find out if x and z both are positive or both are negative and is x> z

Statement (1)
z < x
z and x both could be positive and both could be negative. We dont know anything about y. Insufficient.

Statement (2)
y>0
x and z both are negative but we dont know if x>z or x<z. Insufficient.

Combining (1) & (2)

x and z both are negative, x>z. Sufficient.

Thus, C
OA?

I got A...
I am not sure, but i'll post. may be someone will find a flaw


Main: zy<xy<0

(1)
z<x, so to make zy<xy<0 right y should be positive
because if y will be negative and we will get zy>xy>0
z and x are negative
let z=-5, x=-3
|x-z|+|x|=|-3+5|+|-3|=5
|z|=|-5|=5
SUFF

(2)
y<0
yeah, I know it from (1) but what about z and y?
Insuff

A

4meonly wrote: Main: zy<xy<0

(1)
z<x, so to make zy<xy<0 right y should be positive
because if y will be negative and we will get zy>xy>0
z and x are negative
let z=-5, x=-3
|x-z|+|x|=|-3+5|+|-3|=5
|z|=|-5|=5
SUFF

You cant assume that y is positive,

what if y is negative, z and x are positive.

z<x
let x = 4
z = 2
|x-z|+|x|= lzl

l4-2l + l4l = l2l
2+4 is not equal to 2


Therefore, statement I is insufficient, because y could be positive as well as negative. We just know x>z.

Hope its clear. Let me know if you still have any doubts.

Morgoth wrote: You cant assume that y is positive,

what if y is negative, z and x are positive.
z<x
let x = 4
z = 2
|x-z|+|x|= lzl
l4-2l + l4l = l2l
2+4 is not equal to 2
Therefore, statement I is insufficient, because y could be positive as well as negative. We just know x>z.
Hope its clear. Let me know if you still have any doubts.

But this will not satisfy main statement - zy<xy<0 if y<0

will be xy<zy but we have zy<xy<0, threfore y<0 from the 1st stem


What is OA?

I made the most terrible mistake again.

I apologize to you "4meonly" I tried to reason with you, presented you with counter argument when it was in fact wrong and you were absolutely correct in your reasoning.

Here is the answer which I think should be correct.

Statement (1)
z<x
if zy < xy <0 , y has to be positive.

Therefore x and z are negative. all the negative numbers satisfy the above equation. Sufficient.

Statement (2)
y>0

if y is greater than 0, then z and x are negative, all negative numbers satisfy the above equation. Sufficient.

Thus, D.


I think this should be the answer. Let me know if anybody thinks otherwise.

Thanks "4meonly" for bringing this question, it has really helped me clear some doubts of my own.

OA?

Morgoth,
no problem. Your posts are helpful for me, too

I agree with you that answer should be D

Can anybody confirm that answer is D?

https://www.beatthegmat.com/absolute-val ... 21744.html[/url]

zy < xy < 0

means that zy and xy have both two numbers with different signs. ( +-) or (-+)

and you will see that they both have "y" in common

SO:

A ) If Y>0 (X and Z) < 0 and X > Z so that we can have zy < xy < 0

B ) if Y<0 (X and Z) > 0 and X < Z so that we can have zy < xy < 0

Lets go back to statements:

1 ) z < x

This means that x - z > 0 so it can get out the absolute value sign as it is

and we also know that if x > Z they have to be both NEGATIVE NUMBERS ( look at A )

so:


IS |x - z| + |x| = |z| ?

X-Z + ( -X ) = -Z

which is |z|, since Z is negative it needs to be -Z

Sufficient

(2) y > 0

We already discussed this

A ) If Y>0 (X and Z) < 0 and X > Z so that we can have zy < xy < 0

Sufficient

Hence, it is D