Is |x − z − y| > x − z + y?

Is |x−z−y|>x−z+y?

(1) 0<x<z<y
|x−z−y|>x−z+y
Because x is less than z is less than y, |x-z-y| is negative, so:
-(x-z-y)>x-z+y
-x+z+y>x-z+y
z+y>2x-z+y
2z>2x
z>x
z>x which is given to us in the statement.
SUFFICIENT

(2) (x–z–y) is negative
|x−z−y|>x−z+y
-(x-z-y)>x-z+y
-x+z+y>x-z+y
2z>2x
z>x
However, we are not given any more information.
INSUFFICIENT

(A)

Bunuel I have one doubt here
|x−z−y|>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative

x−z−y>x−z+y --(1) and -x+z+y>x−z+y --(2)
solving 1 we y<0
solving second we get x<z

now if we see option A it gives us hint that basically
|x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y?--- so I thought Its sufficient
Now in case of B we already know that mode is negative ... so why cant it be sufficient

Please help me understand what M i missing here

I think the key here is to note that statement one DOES NOT satisfy the condition of y < 0 (the statement says that 0<x<y<z) and thus, we need to only take care of the other case and can ignore the y < 0 case which makes statement 1 sufficient.

In other words, if we solve the |x-z-y|, we can have two cases - one positive and one negative as you specified above.

Now, if we talk about statement 2, it says that (x–z–y) is negative (less than zero). So, this basically means that our expression in the absolute expression (in the question stem is negative) or in other words, |x−z−y| = - (x-z-y). BUT, this does not answer the main question we have in the question stem.

Knowing that the absolute value of an expression is negative does NOT answer the main question at hand which is whether |x−z−y| > x−z+y?

Hope the information helps. All the best to you !!

Official solution from Veritas Prep.

Answer: A

Explanation: For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If \(x - z - y\) is positive, then the question is really “Is \(x - z - y > x - z + y\)?" which can be simplified by adding/subtracting terms to “Is \(y < 0\)” In other words, if you can prove that x-z-y is positive you then only need to determine the sign of y to answer the question. However if \(x - z - y\) is negative then you need to answer this question: “Is \(- (x - z - y) > x - z + y\)?” Simplifying that question by adding/subtracting terms you see that the question is really “Is \(x < z\)?” In statement 1, you learn cleverly that indeed \((x – z – y)\) is negative because if you subtract progressively bigger positive numbers from an original positive number, the result will always be negative. Given that \((x – z – y)\) is negative then you need to answer the question “Is \(x < z?\)” and this is expressly given in the statement. Statement (1) is sufficient. Statement (2) is tempting you with information that you already know from statement (1). By itself, it is not sufficient because you do not know if \(x < z\). Answer is (A).

Let's pick a number:
0<x<z<y
X=1; Z=2; y=10
Is |x−z−y|>x−z+y?
|1-2-10|>1-2+10
or,11>11
NO
Again,
Let's pick a number:
0<x<z<y
X=1; Z=2; y=3
Is |x−z−y|>x−z+y?
|1-2-3|>1-2+3
or,4>2
Yes
Thus, Statement 1 is insufficient, I think.

Hello

Please check the highlighted part. 1-2+10 will be 9, not 11.

For this type of difficult inequality problem, you want to do as much algebraic manipulation as possible. First you should consider the two possible cases for the absolute value in the question (use the two case approach), and then simplify the question. If x - z - y is positive, then the question is really “Is x - z - y > x - z + y?" which can be simplified to “Is y < 0?” by adding/subtracting terms. In other words, if you can prove that x-z-y is positive you then only need to determine the sign of y to answer the question. However if x - z - y is negative then you need to answer this question: “Is - (x - z - y) > x - z + y?” Simplifying that question by adding/subtracting terms, you see that the question is really “Is x < z?”

When you simplify |x − z − y| > x − z + y? You get two possibilities i.e. “Is y<0” or “Is x<z?”

St1 is in confirmation with Y is not less than zero so we have a definite "No" hence st 1 is sufficient.

St 2 leads us nowhere hence insufficient

Option A is the correct answer

First check this blog post: https://anaprep.com/algebra-the-why-beh ... questions/

Question: Is |x − z − y| > x − z + y?

What is |x − z − y|? How do we evaluate it? We need to know whether x − z − y is positive or negative.

(1) 0 < x < z < y

If x is to the left of z and to the left of y on the number line, (z + y) is much greater than x so x - (z + y) is negative. Hence x - z - y is negative.
|x − z − y| = -(x - z - y) = -x + z + y

Question: Is -x + z + y > x − z + y?
Is z > x?
We are given it is hence sufficient.


(2) (x – z – y) is negative

Then |x − z − y| = -(x - z - y) = -x + z + y

Question: Is -x + z + y > x − z + y?
Is z > x?

We don't know. All we know is that (x – z – y) is negative. z may be greater than x as in statement 1 above or it may be less than x as in x = 2, z = 1 and y = 10. This also satisfies (x – z – y) is negative.
Hence statement 2 alone is not sufficient.

Answer (A)